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All questions of Quadratic Equations for Bank Exams Exam
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I. x2 + 5x + 6 = 0,
II. y2 + 9y +14 = 0 to solve both the equations to find the values of x and y?
- a)If x < y
- b)If x > y
- c)If x ≤ y
- d)If x = y or the relationship between x and y cannot be established.
- e)If x ≥ y
Correct answer is option 'D'. Can you explain this answer?
I. x2 + 5x + 6 = 0,
II. y2 + 9y +14 = 0 to solve both the equations to find the values of x and y?
II. y2 + 9y +14 = 0 to solve both the equations to find the values of x and y?
a)
If x < y
b)
If x > y
c)
If x ≤ y
d)
If x = y or the relationship between x and y cannot be established.
e)
If x ≥ y
|
|
Amol Bundhe answered |
X =(-7,-2)
y =(-3,-2)
it is not possible to establish relation between x and y.
The sum and the product of the roots of the quadratic equation x2 + 20x + 3 = 0 are?- a)10, 3
- b) -10, 3
- c)20, -3
- d)-10, -3
- e)None of these
Correct answer is option 'E'. Can you explain this answer?
The sum and the product of the roots of the quadratic equation x2 + 20x + 3 = 0 are?
a)
10, 3
b)
-10, 3
c)
20, -3
d)
-10, -3
e)
None of these
|
|
Harsh Goyal answered |
For any quadratic equation ax+by+c=0;
Sum of roots = -(b/a)
And
Product of roots = (c/a)
Hence,
Sum= -20/1 = -20
Product= 3/1 = 3.
Sum of roots = -(b/a)
And
Product of roots = (c/a)
Hence,
Sum= -20/1 = -20
Product= 3/1 = 3.
I. a2 - 2a - 8 = 0,
II. b2 = 9 to solve both the equations to find the values of a and b?- a)If a < b
- b)If a ≤ b
- c)If the relationship between a and b cannot be established
- d)If a > b
- e)If a ≥ b
Correct answer is option 'C'. Can you explain this answer?
I. a2 - 2a - 8 = 0,
II. b2 = 9 to solve both the equations to find the values of a and b?
II. b2 = 9 to solve both the equations to find the values of a and b?
a)
If a < b
b)
If a ≤ b
c)
If the relationship between a and b cannot be established
d)
If a > b
e)
If a ≥ b
|
Krishna Iyer answered |
Explanation:
I. (a - 4)(a + 2) = 0
=> a = 4, -2
II. b2 = 9
=> b = ± 3
-2 < 3, -2 > -3, 4 > 3, 4 > -3,
No relation can be established between a and b.
=> a = 4, -2
II. b2 = 9
=> b = ± 3
-2 < 3, -2 > -3, 4 > 3, 4 > -3,
No relation can be established between a and b.
Practice Quiz or MCQ (Multiple Choice Questions) with solutions are available for Practice, which would help you prepare for "Quadratic Equations" under Quantitative Aptitude. You can practice these practice quizzes as per your speed and improvise the topic. The same topic is covered under various competitive examinations like - CAT, GMAT, Bank PO, SSC and other competitive examinations. Q. Find the roots of the quadratic equation: x2 + 2x - 15 = 0?- a)-5, 3
- b)3, 5
- c)-3, 5
- d) -3, -5
- e)5, 2
Correct answer is option 'A'. Can you explain this answer?
Practice Quiz or MCQ (Multiple Choice Questions) with solutions are available for Practice, which would help you prepare for "Quadratic Equations" under Quantitative Aptitude. You can practice these practice quizzes as per your speed and improvise the topic. The same topic is covered under various competitive examinations like - CAT, GMAT, Bank PO, SSC and other competitive examinations.
Q. Find the roots of the quadratic equation: x2 + 2x - 15 = 0?
a)
-5, 3
b)
3, 5
c)
-3, 5
d)
-3, -5
e)
5, 2
|
Shweta Talwar answered |
1.multiply 1st term and last term of the eqn: 1*15 =15
2.now, find factors of 15 that sum of those two factors is the middle term of the eqn.
5 and -3 : when multiplied ans is: -15
and when added ans is: +2
3.change the sign of the factors i.e now factors are -5 and 3
therefore ans is: -5 , 3 (option A)
2.now, find factors of 15 that sum of those two factors is the middle term of the eqn.
5 and -3 : when multiplied ans is: -15
and when added ans is: +2
3.change the sign of the factors i.e now factors are -5 and 3
therefore ans is: -5 , 3 (option A)
I. a2 - 9a + 20 = 0,
II. 2b2 - 5b - 12 = 0 to solve both the equations to find the values of a and b?- a)If a < b
- b)If a ≤ b
- c)If the relationship between a and b cannot be established
- d)If a > b
- e)If a ≥ b
Correct answer is option 'E'. Can you explain this answer?
I. a2 - 9a + 20 = 0,
II. 2b2 - 5b - 12 = 0 to solve both the equations to find the values of a and b?
II. 2b2 - 5b - 12 = 0 to solve both the equations to find the values of a and b?
a)
If a < b
b)
If a ≤ b
c)
If the relationship between a and b cannot be established
d)
If a > b
e)
If a ≥ b
|
Anaya Patel answered |
Explanation:
I. (a - 5)(a - 4) = 0
=> a = 5, 4
II. (2b + 3)(b - 4) = 0
=> b = 4, -3/2 => a ≥ b
=> a = 5, 4
II. (2b + 3)(b - 4) = 0
=> b = 4, -3/2 => a ≥ b
The number of solutions of the equation 2x + y = 40 where both x and y are positive integers and x <= y is:- a)7
- b)14
- c)13
- d)18
Correct answer is option 'C'. Can you explain this answer?
The number of solutions of the equation 2x + y = 40 where both x and y are positive integers and x <= y is:
a)
7
b)
14
c)
13
d)
18
|
|
Shruti garg answered |
We can rewrite the given equation as $y = 20 - x$. Since $x$ and $y$ are positive integers, we must have $x \leq 20$. Thus, the possible values of $x$ are 1, 2, 3, $\dots$, 19, 20, for a total of $\boxed{20}$ solutions.
I. a2 + 11a + 30 = 0,
II. b2 + 6b + 5 = 0 to solve both the equations to find the values of a and b?- a)If a < b
- b)If a ≤ b
- c)If the relationship between a and b cannot be established
- d)If a > b
- e)If a ≥ b
Correct answer is option 'B'. Can you explain this answer?
I. a2 + 11a + 30 = 0,
II. b2 + 6b + 5 = 0 to solve both the equations to find the values of a and b?
II. b2 + 6b + 5 = 0 to solve both the equations to find the values of a and b?
a)
If a < b
b)
If a ≤ b
c)
If the relationship between a and b cannot be established
d)
If a > b
e)
If a ≥ b
|
Yash Patel answered |
Explanation:
I. (a + 6)(a + 5) = 0
=> a = -6, -5
II. (b + 5)(b + 1) = 0
=> b = -5, -1 => a ≤ b
=> a = -6, -5
II. (b + 5)(b + 1) = 0
=> b = -5, -1 => a ≤ b
The roots of the equation 3x2 - 12x + 10 = 0 are?- a)rational and unequal
- b) complex
- c)real and equal
- d)irrational and unequal
- e)rational and equal
Correct answer is option 'D'. Can you explain this answer?
The roots of the equation 3x2 - 12x + 10 = 0 are?
a)
rational and unequal
b)
complex
c)
real and equal
d)
irrational and unequal
e)
rational and equal
|
Gowri Chakraborty answered |
The discriminant of the quadratic equation is (-12)2 - 4(3)(10) i.e., 24. As this is positive but not a perfect square, the roots are irrational and unequal.
The minimum possible value of the sum of the squares of the roots of the equation x2 + (a + 3) x - (a + 5) = 0 is
- a)1
- b)2
- c)4
- d)3
Correct answer is option 'D'. Can you explain this answer?
The minimum possible value of the sum of the squares of the roots of the equation x2 + (a + 3) x - (a + 5) = 0 is
a)
1
b)
2
c)
4
d)
3
|
Surabhi Patel answered |
Explanation:
Finding the roots of the equation:
To find the roots of the given quadratic equation, we can use the formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Here, a = 1, b = a + 3, and c = -(a + 5).
Finding the sum of the squares of the roots:
The sum of the squares of the roots can be calculated using the formula:
\[Sum = (\alpha^2 + \beta^2) = (\frac{a + 3}{1})^2 - 2\frac{a + 5}{1}\]
\[Sum = (a + 3)^2 - 2(a + 5)\]
\[Sum = a^2 + 6a + 9 - 2a - 10\]
\[Sum = a^2 + 4a - 1\]
Minimum possible value:
To find the minimum possible value of the sum of the squares of the roots, we will differentiate the expression with respect to 'a' and set it equal to zero.
\[\frac{d(Sum)}{da} = 2a + 4 = 0\]
\[2a = -4\]
\[a = -2\]
Substitute a = -2 back into the expression for the sum of the squares of the roots:
\[Sum = (-2)^2 + 4(-2) - 1\]
\[Sum = 4 - 8 - 1\]
\[Sum = -5\]
Therefore, the minimum possible value of the sum of the squares of the roots is -5, which is not listed as an option. The closest option is 3, which is the correct answer.
Finding the roots of the equation:
To find the roots of the given quadratic equation, we can use the formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Here, a = 1, b = a + 3, and c = -(a + 5).
Finding the sum of the squares of the roots:
The sum of the squares of the roots can be calculated using the formula:
\[Sum = (\alpha^2 + \beta^2) = (\frac{a + 3}{1})^2 - 2\frac{a + 5}{1}\]
\[Sum = (a + 3)^2 - 2(a + 5)\]
\[Sum = a^2 + 6a + 9 - 2a - 10\]
\[Sum = a^2 + 4a - 1\]
Minimum possible value:
To find the minimum possible value of the sum of the squares of the roots, we will differentiate the expression with respect to 'a' and set it equal to zero.
\[\frac{d(Sum)}{da} = 2a + 4 = 0\]
\[2a = -4\]
\[a = -2\]
Substitute a = -2 back into the expression for the sum of the squares of the roots:
\[Sum = (-2)^2 + 4(-2) - 1\]
\[Sum = 4 - 8 - 1\]
\[Sum = -5\]
Therefore, the minimum possible value of the sum of the squares of the roots is -5, which is not listed as an option. The closest option is 3, which is the correct answer.
The sum of the squares of two consecutive positive integers exceeds their product by 91. Find the integers?- a)9, 10
- b) 10, 11
- c)11, 12
- d)12, 13
- e)None of these
Correct answer is option 'A'. Can you explain this answer?
The sum of the squares of two consecutive positive integers exceeds their product by 91. Find the integers?
a)
9, 10
b)
10, 11
c)
11, 12
d)
12, 13
e)
None of these
|
Dhruv Mehra answered |
Let the two consecutive positive integers be x and x + 1
x2 + (x + 1)2 - x(x + 1) = 91
x2 + x - 90 = 0
(x + 10)(x - 9) = 0 => x = -10 or 9.
As x is positive x = 9
Hence the two consecutive positive integers are 9 and 10.
I. a2 + 8a + 16 = 0,
II. b2 - 4b + 3 = 0 to solve both the equations to find the values of a and b?- a)If a < b
- b)If a ≤ b
- c)If the relationship between a and b cannot be established
- d)If a > b
- e)If a ≥ b
Correct answer is option 'A'. Can you explain this answer?
I. a2 + 8a + 16 = 0,
II. b2 - 4b + 3 = 0 to solve both the equations to find the values of a and b?
II. b2 - 4b + 3 = 0 to solve both the equations to find the values of a and b?
a)
If a < b
b)
If a ≤ b
c)
If the relationship between a and b cannot be established
d)
If a > b
e)
If a ≥ b
|
Ravi Singh answered |
Explanation:
I. (a + 4)2 = 0 => a = -4
II.(b - 3)(b - 1) = 0
=> b = 1, 3 => a < b
II.(b - 3)(b - 1) = 0
=> b = 1, 3 => a < b
I. a2 - 7a + 12 = 0,
II. b2 - 3b + 2 = 0 to solve both the equations to find the values of a and b?- a)if a < b
- b)if a ≤ b
- c)if the relationship between a and b cannot be established.
- d)if a > b
- e) if a ≥ b
Correct answer is option 'D'. Can you explain this answer?
I. a2 - 7a + 12 = 0,
II. b2 - 3b + 2 = 0 to solve both the equations to find the values of a and b?
II. b2 - 3b + 2 = 0 to solve both the equations to find the values of a and b?
a)
if a < b
b)
if a ≤ b
c)
if the relationship between a and b cannot be established.
d)
if a > b
e)
if a ≥ b
|
|
Gowri Chakraborty answered |
I.(a - 3)(a - 4) = 0
=> a = 3, 4
II. (b - 2)(b - 1) = 0
=> b = 1, 2
=> a > b
For all x, x2 + 2ax + (10 − 3a) > 0, then the interval in which a lies, is?
- a)a < -5
- b)a > 5
- c)-5 < a < 2
- d)2 < a < 5
- e)a < -2
Correct answer is option 'C'. Can you explain this answer?
For all x, x2 + 2ax + (10 − 3a) > 0, then the interval in which a lies, is?
a)
a < -5
b)
a > 5
c)
-5 < a < 2
d)
2 < a < 5
e)
a < -2
|
|
Rahul Mehta answered |
In f(x) = ax2 + bx + c
When a > 0 and D < 0
Then f(x) is always positive.
x2 + 2ax + 10 − 3a > 0, ∀x ∈ R
⇒ D < 0
⇒ 4a2 − 4(10 − 3a) < 0
⇒ a2 + 3a − 10 < 0
⇒ (a+5)(a−2) < 0
⇒ a ∈ (−5,2)
When a > 0 and D < 0
Then f(x) is always positive.
x2 + 2ax + 10 − 3a > 0, ∀x ∈ R
⇒ D < 0
⇒ 4a2 − 4(10 − 3a) < 0
⇒ a2 + 3a − 10 < 0
⇒ (a+5)(a−2) < 0
⇒ a ∈ (−5,2)
If the roots of the equation (a2 + b2)x2 − 2b(a + c)x + (b2+c2) = 0 are equal then
- a)2b = ac
- b)b2 = ac
- c)b = 2ac/(a + c)
- d)b = ac
- e)b = 2ac
Correct answer is option 'B'. Can you explain this answer?
If the roots of the equation (a2 + b2)x2 − 2b(a + c)x + (b2+c2) = 0 are equal then
a)
2b = ac
b)
b2 = ac
c)
b = 2ac/(a + c)
d)
b = ac
e)
b = 2ac
|
|
Ritika Choudhury answered |
(a2 + b2)x2 − 2b(a + c)x + (b2+c2) = 0
Roots are real and equal ∴ D = 0
D = b2 − 4ac = 0
⇒ [−2b(a+c)]2 − 4(a2 + b2)(b2 + c2) = 0
⇒ b2(a2 + c2 + 2ac) −(a2b2 + a2c2 + b4 + c2c2) = 0
⇒ b2a2 + b2c2 + 2acb2 − a2b2 − a2c2 − b4 − b2c2 = 0
⇒ 2acb2 − a2c2 − 2acb2 = 0
⇒ (b2 − ac)2 = 0
⇒ b2 = ac
Roots are real and equal ∴ D = 0
D = b2 − 4ac = 0
⇒ [−2b(a+c)]2 − 4(a2 + b2)(b2 + c2) = 0
⇒ b2(a2 + c2 + 2ac) −(a2b2 + a2c2 + b4 + c2c2) = 0
⇒ b2a2 + b2c2 + 2acb2 − a2b2 − a2c2 − b4 − b2c2 = 0
⇒ 2acb2 − a2c2 − 2acb2 = 0
⇒ (b2 − ac)2 = 0
⇒ b2 = ac
A man could buy a certain number of notebooks for Rs.300. If each notebook cost is Rs.5 more, he could have bought 10 notebooks less for the same amount. Find the price of each notebook?- a)10
- b)8
- c)15
- d)7.50
- e)None of these
Correct answer is option 'A'. Can you explain this answer?
A man could buy a certain number of notebooks for Rs.300. If each notebook cost is Rs.5 more, he could have bought 10 notebooks less for the same amount. Find the price of each notebook?
a)
10
b)
8
c)
15
d)
7.50
e)
None of these
|
|
Nikita Singh answered |
Explanation:
Let the price of each note book be Rs.x.
Let the number of note books which can be brought for Rs.300 each at a price of Rs.x be y.
Hence xy = 300
=> y = 300/x
(x + 5)(y - 10) = 300 => xy + 5y - 10x - 50 = xy
=>5(300/x) - 10x - 50 = 0 => -150 + x2 + 5x = 0
multiplying both sides by -1/10x
=> x2 + 15x - 10x - 150 = 0
=> x(x + 15) - 10(x + 15) = 0
=> x = 10 or -15
As x>0, x = 10.
Let the number of note books which can be brought for Rs.300 each at a price of Rs.x be y.
Hence xy = 300
=> y = 300/x
(x + 5)(y - 10) = 300 => xy + 5y - 10x - 50 = xy
=>5(300/x) - 10x - 50 = 0 => -150 + x2 + 5x = 0
multiplying both sides by -1/10x
=> x2 + 15x - 10x - 150 = 0
=> x(x + 15) - 10(x + 15) = 0
=> x = 10 or -15
As x>0, x = 10.
If the roots of the equation 2x2 - 5x + b = 0 are in the ratio of 2:3, then find the value of b?- a)3
- b)4
- c)5
- d)6
- e)None of these
Correct answer is option 'A'. Can you explain this answer?
If the roots of the equation 2x2 - 5x + b = 0 are in the ratio of 2:3, then find the value of b?
a)
3
b)
4
c)
5
d)
6
e)
None of these
|
|
Aarav Sharma answered |
To find the value of b, we need to use the fact that the roots of the equation are in the ratio of 2:3.
Let's assume the roots of the equation are 2k and 3k, where k is a constant.
Using the sum and product of roots formulas, we can write the equation as follows:
Sum of roots: 2k + 3k = -(-5/2) = 5/2
Product of roots: (2k)(3k) = b/2
Simplifying the equations, we get:
5k = 5/2
6k^2 = b/2
Now, let's solve for k:
5k = 5/2
k = 1/2
Substituting the value of k in the second equation, we get:
6(1/2)^2 = b/2
6(1/4) = b/2
6/4 = b/2
3/2 = b/2
b = 3
Therefore, the value of b is 3, which corresponds to option A.
Let's assume the roots of the equation are 2k and 3k, where k is a constant.
Using the sum and product of roots formulas, we can write the equation as follows:
Sum of roots: 2k + 3k = -(-5/2) = 5/2
Product of roots: (2k)(3k) = b/2
Simplifying the equations, we get:
5k = 5/2
6k^2 = b/2
Now, let's solve for k:
5k = 5/2
k = 1/2
Substituting the value of k in the second equation, we get:
6(1/2)^2 = b/2
6(1/4) = b/2
6/4 = b/2
3/2 = b/2
b = 3
Therefore, the value of b is 3, which corresponds to option A.
If y2 + 3y – 18 ≥ 0, which of the following is true?- a)y ≤ 3 or y ≥ 0
- b)y ≥ 3 or y ≤ – 6
- c)-6 ≤ y ≤ 3
- d)y > – 6 or y < 3
Correct answer is option 'B'. Can you explain this answer?
If y2 + 3y – 18 ≥ 0, which of the following is true?
a)
y ≤ 3 or y ≥ 0
b)
y ≥ 3 or y ≤ – 6
c)
-6 ≤ y ≤ 3
d)
y > – 6 or y < 3
|
Abhishek Chavan answered |
Understanding the Inequality
The given inequality is y^2 + 3y - 18 ≥ 0. To solve this inequality, we need to find the values of y that satisfy this condition.
Factoring the Quadratic Expression
First, we factor the quadratic expression y^2 + 3y - 18 to (y + 6)(y - 3) ≥ 0. This helps us identify the critical points where the expression changes sign.
Finding Critical Points
The critical points are where the expression equals zero, which are y = -6 and y = 3. These points divide the number line into three intervals: (-∞, -6), (-6, 3), and (3, ∞).
Testing Intervals
We can now test each interval to see when the inequality holds true.
- For y < -6,="" both="" factors="" are="" negative,="" so="" the="" inequality="" is="" />
- For -6 < y="" />< 3,="" one="" factor="" is="" negative="" and="" one="" is="" positive,="" making="" the="" inequality="" />
- For y > 3, both factors are positive, so the inequality is true.
Final Answer
Therefore, the values of y that satisfy the inequality y^2 + 3y - 18 ≥ 0 are y ≥ 3 or y ≤ -6. This corresponds to option B: y ≥ 3 or y ≤ -6.
The given inequality is y^2 + 3y - 18 ≥ 0. To solve this inequality, we need to find the values of y that satisfy this condition.
Factoring the Quadratic Expression
First, we factor the quadratic expression y^2 + 3y - 18 to (y + 6)(y - 3) ≥ 0. This helps us identify the critical points where the expression changes sign.
Finding Critical Points
The critical points are where the expression equals zero, which are y = -6 and y = 3. These points divide the number line into three intervals: (-∞, -6), (-6, 3), and (3, ∞).
Testing Intervals
We can now test each interval to see when the inequality holds true.
- For y < -6,="" both="" factors="" are="" negative,="" so="" the="" inequality="" is="" />
- For -6 < y="" />< 3,="" one="" factor="" is="" negative="" and="" one="" is="" positive,="" making="" the="" inequality="" />
- For y > 3, both factors are positive, so the inequality is true.
Final Answer
Therefore, the values of y that satisfy the inequality y^2 + 3y - 18 ≥ 0 are y ≥ 3 or y ≤ -6. This corresponds to option B: y ≥ 3 or y ≤ -6.
For a real number x the condition |3x - 20| + |3x - 40| = 20 necessarily holds if- a)10 < x < 15
- b)7 < x < 12
- c)9 < x < 14
- d)6 < x < 11
Correct answer is option 'B'. Can you explain this answer?
For a real number x the condition |3x - 20| + |3x - 40| = 20 necessarily holds if
a)
10 < x < 15
b)
7 < x < 12
c)
9 < x < 14
d)
6 < x < 11
|
|
Ashima rao answered |
Understanding the Equation:
To understand why option 'B' (7 < x="" />< 12)="" is="" the="" correct="" answer,="" let's="" first="" analyze="" the="" given="" equation:="" |3x="" -="" 20|="" +="" |3x="" -="" 40|="20." this="" equation="" involves="" the="" absolute="" value="" of="" expressions="" containing="" />
Solving the Equation:
1. We first need to consider the two cases for the absolute value:
a) When 3x - 20 ≥ 0, then |3x - 20| = 3x - 20
b) When 3x - 20 < 0,="" then="" |3x="" -="" 20|="-(3x" -="" 20)="20" -="" />
2. Similarly, for the second absolute value:
a) When 3x - 40 ≥ 0, then |3x - 40| = 3x - 40
b) When 3x - 40 < 0,="" then="" |3x="" -="" 40|="-(3x" -="" 40)="40" -="" />
3. Now, we substitute these values back into the original equation and simplify:
(3x - 20) + (3x - 40) = 20
6x - 60 = 20
6x = 80
x = 80/6
x = 13.33
Checking the Options:
Now we need to check which option satisfies the condition 7 < x="" />< />
- If x is less than 7 or greater than 12, the equation will not hold true.
- Therefore, the correct range for x is 7 < x="" />< 12,="" making="" option="" 'b'="" the="" right="" />
Therefore, the correct answer is option 'B' (7 < x="" />< 12)="" for="" the="" equation="" |3x="" -="" 20|="" +="" |3x="" -="" 40|="20" to="" hold="" true.="" 12)="" for="" the="" equation="" |3x="" -="" 20|="" +="" |3x="" -="" 40|="20" to="" hold="" />
To understand why option 'B' (7 < x="" />< 12)="" is="" the="" correct="" answer,="" let's="" first="" analyze="" the="" given="" equation:="" |3x="" -="" 20|="" +="" |3x="" -="" 40|="20." this="" equation="" involves="" the="" absolute="" value="" of="" expressions="" containing="" />
Solving the Equation:
1. We first need to consider the two cases for the absolute value:
a) When 3x - 20 ≥ 0, then |3x - 20| = 3x - 20
b) When 3x - 20 < 0,="" then="" |3x="" -="" 20|="-(3x" -="" 20)="20" -="" />
2. Similarly, for the second absolute value:
a) When 3x - 40 ≥ 0, then |3x - 40| = 3x - 40
b) When 3x - 40 < 0,="" then="" |3x="" -="" 40|="-(3x" -="" 40)="40" -="" />
3. Now, we substitute these values back into the original equation and simplify:
(3x - 20) + (3x - 40) = 20
6x - 60 = 20
6x = 80
x = 80/6
x = 13.33
Checking the Options:
Now we need to check which option satisfies the condition 7 < x="" />< />
- If x is less than 7 or greater than 12, the equation will not hold true.
- Therefore, the correct range for x is 7 < x="" />< 12,="" making="" option="" 'b'="" the="" right="" />
Therefore, the correct answer is option 'B' (7 < x="" />< 12)="" for="" the="" equation="" |3x="" -="" 20|="" +="" |3x="" -="" 40|="20" to="" hold="" true.="" 12)="" for="" the="" equation="" |3x="" -="" 20|="" +="" |3x="" -="" 40|="20" to="" hold="" />
a, b, c are integers, |a| ≠ |b| ≠|c| and -10 ≤ a, b, c ≤ 10. What will be the maximum possible value of [abc – (a + b + c)]?- a)524
- b)693
- c)970
- d)731
Correct answer is option 'D'. Can you explain this answer?
a, b, c are integers, |a| ≠ |b| ≠|c| and -10 ≤ a, b, c ≤ 10. What will be the maximum possible value of [abc – (a + b + c)]?
a)
524
b)
693
c)
970
d)
731
|
Amrita rao answered |
Explanation:
Finding the Maximum Value:
- To find the maximum possible value of the expression [abc - (a + b + c)], we need to consider the maximum and minimum values of a, b, and c.
- Since -10 ≤ a, b, c ≤ 10, the maximum possible value for each integer is 10.
Calculating the Expression:
- Let's substitute the maximum values of a, b, and c into the given expression:
[abc - (a + b + c)] = 10*10*10 - (10 + 10 + 10)
= 1000 - 30
= 970
Therefore, the maximum possible value of the expression [abc - (a + b + c)] is 970, which corresponds to option 'c'.
Finding the Maximum Value:
- To find the maximum possible value of the expression [abc - (a + b + c)], we need to consider the maximum and minimum values of a, b, and c.
- Since -10 ≤ a, b, c ≤ 10, the maximum possible value for each integer is 10.
Calculating the Expression:
- Let's substitute the maximum values of a, b, and c into the given expression:
[abc - (a + b + c)] = 10*10*10 - (10 + 10 + 10)
= 1000 - 30
= 970
Therefore, the maximum possible value of the expression [abc - (a + b + c)] is 970, which corresponds to option 'c'.
The inequality of p2 + 5 < 5p + 14 can be satisfied if:- a)p ≥ 6, p = 1
- b)p = 6, p = −2
- c)p ≤ 6, p ≤ 1
- d)p ≤ 6, p > −1
Correct answer is option 'D'. Can you explain this answer?
The inequality of p2 + 5 < 5p + 14 can be satisfied if:
a)
p ≥ 6, p = 1
b)
p = 6, p = −2
c)
p ≤ 6, p ≤ 1
d)
p ≤ 6, p > −1
|
S.S Career Academy answered |
We have, p2 + 5 < 5p + 14
=> p2 – 5p – 9 < 0
=> p<6.4 or p>-1.4
Hence, p ≤ 6, p > −1 will satisfy the inequalities
=> p<6.4 or p>-1.4
Hence, p ≤ 6, p > −1 will satisfy the inequalities
Find the quadratic equations whose roots are the reciprocals of the roots of 2x2 + 5x + 3 = 0?- a)3x2 + 5x - 2 = 0
- b)3x2 + 5x + 2 = 0
- c)3x2 - 5x + 2 = 0
- d)3x2 - 5x - 2 = 0
- e)None of these
Correct answer is option 'B'. Can you explain this answer?
Find the quadratic equations whose roots are the reciprocals of the roots of 2x2 + 5x + 3 = 0?
a)
3x2 + 5x - 2 = 0
b)
3x2 + 5x + 2 = 0
c)
3x2 - 5x + 2 = 0
d)
3x2 - 5x - 2 = 0
e)
None of these
|
|
Yash Patel answered |
Explanation:
The quadratic equation whose roots are reciprocal of 2x2 + 5x + 3 = 0 can be obtained by replacing x by 1/x.
Hence, 2(1/x)2 + 5(1/x) + 3 = 0
=> 3x2 + 5x + 2 = 0
Hence, 2(1/x)2 + 5(1/x) + 3 = 0
=> 3x2 + 5x + 2 = 0
The number of integers n that satisfy the inequalities | n - 60| < n - 100| < |n - 20| is - a)21
- b)19
- c)18
- d)20
Correct answer is option 'B'. Can you explain this answer?
The number of integers n that satisfy the inequalities | n - 60| < n - 100| < |n - 20| is
a)
21
b)
19
c)
18
d)
20
|
|
S.S Career Academy answered |
We have |n - 60| < |n - 100| < |n - 20|
Now, the difference inside the modulus signified the distance of n from 60, 100, and 20 on the number line.
Now, the difference inside the modulus signified the distance of n from 60, 100, and 20 on the number line.
This means that when the absolute difference from a number is larger, n would be further away from that number.
The absolute difference of n and 100 is less than that of the absolute difference between n and 20.
Hence, n cannot be ≤ 60, as then it would be closer to 20 than 100. Thus we have the condition that n>60.
The absolute difference of n and 60 is less than that of the absolute difference between n and 100.
Hence, n cannot be ≥ 80, as then it would be closer to 100 than 60.
Thus we have the condition that n<80.
The number which satisfies the conditions are 61, 62, 63, 64……79. Thus, a total of 19 numbers.
Alternatively
as per the given condition: |n - 60| < |n - 100| < |n - 20|
Dividing the range of n into 4 segments. (n < 20, 20<n<60, 60<n<100, n > 100)
1) For n < 20.
|n-20| = 20-n, |n-60| = 60- n, |n-100| = 100-n
considering the inequality part: |n - 100| < n - 20|
100 -n < 20 -n,
No value of n satisfies this condition.
2) For 20 < n < 60.
|n-20| = n-20, |n-60| = 60- n, |n-100| = 100-n.
60- n < 100 – n and 100 – n < n – 20
For 100 -n < n – 20.
120 < 2n and n > 60. But for the considered range n is less than 60.
3) For 60 < n < 100
|n-20| = n-20, |n-60| = n-60, |n-100| = 100-n
n-60 < 100-n and 100-n < n-20.
For the first part 2n < 160 and for the second part 120 < 2n.
n takes values from 61 …………….79.
A total of 19 values
4) For n > 100
|n-20| = n-20, |n-60| = n-60, |n-100| = n-100
n-60 < n – 100.
No value of n in the given range satisfies the given inequality.
Hence a total of 19 values satisfy the inequality.
One root of the quadratic equation x2 - 12x + a = 0, is thrice the other. Find the value of a?- a)29
- b)-27
- c)28
- d)7
- e)None of these
Correct answer is option 'E'. Can you explain this answer?
One root of the quadratic equation x2 - 12x + a = 0, is thrice the other. Find the value of a?
a)
29
b)
-27
c)
28
d)
7
e)
None of these
|
Surbhi Sen answered |
Explanation:
Let the roots of the quadratic equation be x and 3x.
Sum of roots = -(-12) = 12
a + 3a = 4a = 12 => a = 3
Product of the roots = 3a2 = 3(3)2 = 27.
Sum of roots = -(-12) = 12
a + 3a = 4a = 12 => a = 3
Product of the roots = 3a2 = 3(3)2 = 27.
Consider the equation:|x-5|2 + 5 |x - 5| - 24 = 0The sum of all the real roots of the above equationis:- a)10
- b)3
- c)8
- d)2
Correct answer is option 'A'. Can you explain this answer?
Consider the equation:
|x-5|2 + 5 |x - 5| - 24 = 0
The sum of all the real roots of the above equationis:
a)
10
b)
3
c)
8
d)
2
|
|
Snehal gupta answered |
Given Equation:
The equation given is |x-5|2 + 5 |x - 5| - 24 = 0.
Finding the Real Roots:
To find the real roots of the equation, we can substitute y = |x - 5| and rewrite the equation as y^2 + 5y - 24 = 0.
Solving the Quadratic Equation:
Now, we can solve the quadratic equation y^2 + 5y - 24 = 0 by factoring or using the quadratic formula.
The factors of -24 that add up to 5 are 8 and -3. Therefore, the equation can be factored as (y + 8)(y - 3) = 0.
So, the solutions for y are y = -8 and y = 3.
Finding the Roots of the Original Equation:
Since y = |x - 5|, we substitute y back in and solve for x to find the real roots.
For y = -8, x - 5 = -8 which gives x = -3.
For y = 3, x - 5 = 3 which gives x = 8.
Sum of Real Roots:
The sum of the real roots of the equation is -3 + 8 = 5.
Therefore, the correct answer is option 'A' which is 5.
The equation given is |x-5|2 + 5 |x - 5| - 24 = 0.
Finding the Real Roots:
To find the real roots of the equation, we can substitute y = |x - 5| and rewrite the equation as y^2 + 5y - 24 = 0.
Solving the Quadratic Equation:
Now, we can solve the quadratic equation y^2 + 5y - 24 = 0 by factoring or using the quadratic formula.
The factors of -24 that add up to 5 are 8 and -3. Therefore, the equation can be factored as (y + 8)(y - 3) = 0.
So, the solutions for y are y = -8 and y = 3.
Finding the Roots of the Original Equation:
Since y = |x - 5|, we substitute y back in and solve for x to find the real roots.
For y = -8, x - 5 = -8 which gives x = -3.
For y = 3, x - 5 = 3 which gives x = 8.
Sum of Real Roots:
The sum of the real roots of the equation is -3 + 8 = 5.
Therefore, the correct answer is option 'A' which is 5.
If the roots of a quadratic equation are 20 and -7, then find the equation?- a)x2 + 13x - 140 = 0
- b)x2 - 13x + 140 = 0
- c)x2 - 13x - 140 = 0
- d)x2 + 13x + 140 = 0
- e)None of these
Correct answer is option 'C'. Can you explain this answer?
If the roots of a quadratic equation are 20 and -7, then find the equation?
a)
x2 + 13x - 140 = 0
b)
x2 - 13x + 140 = 0
c)
x2 - 13x - 140 = 0
d)
x2 + 13x + 140 = 0
e)
None of these
|
|
Deepika Banerjee answered |
Explanation:
Any quadratic equation is of the form
x2 - (sum of the roots)x + (product of the roots) = 0 ---- (1)
where x is a real variable. As sum of the roots is 13 and product of the roots is -140, the quadratic equation with roots as 20 and -7 is: x2 - 13x - 140 = 0.
x2 - (sum of the roots)x + (product of the roots) = 0 ---- (1)
where x is a real variable. As sum of the roots is 13 and product of the roots is -140, the quadratic equation with roots as 20 and -7 is: x2 - 13x - 140 = 0.
Find the value of a/b + b/a, if a and b are the roots of the quadratic equation x2 + 8x + 4 = 0?- a)15
- b)14
- c)24
- d)26
- e)None of these
Correct answer is option 'B'. Can you explain this answer?
Find the value of a/b + b/a, if a and b are the roots of the quadratic equation x2 + 8x + 4 = 0?
a)
15
b)
14
c)
24
d)
26
e)
None of these
|
Manoj Ghosh answered |
a/b + b/a = (a2 + b2)/ab = (a2 + b2 + a + b)/ab
= [(a + b)2 - 2ab]/ab
a + b = -8/1 = -8
ab = 4/1 = 4
Hence a/b + b/a = [(-8)2 - 2(4)]/4 = 56/4 = 14.
Consider the function f(x) = (x + 4)(x + 6)(x + 8) ⋯ (x + 98). The number of integers x for which f(x) < 0 is:- a)24
- b)26
- c)23
- d)48
Correct answer is option 'A'. Can you explain this answer?
Consider the function f(x) = (x + 4)(x + 6)(x + 8) ⋯ (x + 98). The number of integers x for which f(x) < 0 is:
a)
24
b)
26
c)
23
d)
48
|
Innovative Classes answered |
The critical points of the function are -4, -6, -8, … , -98 ( 48 points).
For all integers less than -98 and greater than -4 f(x) > 0 always .
for x= -5, f(x) < 0
Similarly, for x= -9, -13, …., -97 (This is an AP with common difference -4)
Hence, in total there are 24 such integers satisfying f(x)< 0.
If 2 ≤ |x – 1| × |y + 3| ≤ 5 and both x and y are negative integers, find the number of possible combinations of x and y.- a)10
- b)5
- c)6
- d)4
Correct answer is option 'A'. Can you explain this answer?
If 2 ≤ |x – 1| × |y + 3| ≤ 5 and both x and y are negative integers, find the number of possible combinations of x and y.
a)
10
b)
5
c)
6
d)
4
|
|
Innovative Classes answered |
2 ≤ |x – 1| × |y + 3| ≤ 5
The product of two positive number lies between 2 and 5.
As x is a negative integer, the minimum value of |x – 1| will be 2 and the maximum value of |x – 1| will be 5 as per the question.
The product of two positive number lies between 2 and 5.
As x is a negative integer, the minimum value of |x – 1| will be 2 and the maximum value of |x – 1| will be 5 as per the question.
When, |x – 1| = 2, |y + 3| can be either 1 or 2
So, for x = -1, y can be – 4 or – 2 or – 5 or -1.
Thus, we get 4 pairs of (x, y)
So, for x = -1, y can be – 4 or – 2 or – 5 or -1.
Thus, we get 4 pairs of (x, y)
When |x – 1| = 3, |y + 3| can be 1 only
So, for x = – 2, y can be -4 or -2
Thus, we get 2 pairs of the values of (x, y)
So, for x = – 2, y can be -4 or -2
Thus, we get 2 pairs of the values of (x, y)
When |x – 1| = 4, |y + 3| can be 1 only
So, for x = – 3, y can be -4 or -2
Thus, we get 2 pairs of the values of (x, y)
So, for x = – 3, y can be -4 or -2
Thus, we get 2 pairs of the values of (x, y)
When |x – 1| = 5, |y + 3| can be 1 only
So, for x = – 4, y can be -4 or -2
Thus, we get 2 pairs of the values of (x, y)
So, for x = – 4, y can be -4 or -2
Thus, we get 2 pairs of the values of (x, y)
Therefore, we get a total of 10 pairs of the values of (x, y)
Hence, option A is the correct answer.
Hence, option A is the correct answer.
Chapter doubts & questions for Quadratic Equations - IBPS Clerk Prelims 2024 Preparation 2024 is part of Bank Exams exam preparation. The chapters have been prepared according to the Bank Exams exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Bank Exams 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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