Question for CAT Previous Year Questions: Quadratic equations
Try yourself:If x and y are real numbers such that x2 + (x - 2y - 1)2 = - 4y (x + y), then the value x − 2y is
[2023]
Explanation
Since x & y are real numbers, (x + 2y) & (x − 2y − 1) are both real.
A square of a real number is always non-negative.
For this reason, for the equation: (x + 2y)2 + (x - 2y - 1)2 = 0 to be true, both (x + 2y) & (x - 2y - 1) must be equal to 0.
x − 2y − 1 = 0
x − 2y = 1
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Question for CAT Previous Year Questions: Quadratic equations
Try yourself:If then is equal to
[2023]
Explanation
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Question for CAT Previous Year Questions: Quadratic equations
Try yourself: The number of integer solutions of equation 2 |x| (x2 + 1) = 5x2 is
[2023]
Correct Answer : 3
Explanation
2 |x| (x2 + 1) = 5x2
Let |x| = k
2k (k2 + 1) = 5k2
Either k = 0; or 2(k2 + 1) = 5k
2k2 – 5k + 2 = 0
2k2 – 4k – k + 2 = 0
2k(k – 2) –1(k – 2) = 0
(2k – 1)(k – 2) = 0
k = 0.5 or k = 2
Therefore, k which is |x|, can take the values 0, 0.5 or 2
So, x can take the values 0, -0.5, 0.5, -2, 2
Since we are looking for integral solutions, x can only take the values 0, 0.5 or 2
Therefore, there are only 3 integral solutions to 2 |x| (x2 + 1) = 5x2.
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Question for CAT Previous Year Questions: Quadratic equations
Try yourself:Let α and β be the two distinct roots of the equation 2x2 - 6x + k =0, such that (α+β) and αβ are the distinct roots of the equation x2 + px + p = 0. Then, the value of 8 (k - p) is
[2023]
Correct Answer : 6
Explanation
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Question for CAT Previous Year Questions: Quadratic equations
Try yourself:The equation x3 + (2r+1)x2 + (4r − 1)x + 2 = 0 has -2 as one of the roots. If the other two roots are real, then the minimum possible non-negative integer value of r is
[2023]
Correct Answer : 2
Explanation
x3 + (2r + 1)x2 + (4r − 1)x + 2 = 0
Since -2 is one of the roots, the cubic equation can be factored as…
(x + 2)(x2 + (2x − 1)x + 1) = 0
Since the other two roots are real, (x2 + (2x − 1)x + 1) = 0 has two real roots.
That is the discriminant of (x2 + (2x − 1)x + 1) = 0 is non-negative.
Therefore the minimum possible non-negative integral value of r is 2.
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Question for CAT Previous Year Questions: Quadratic equations
Try yourself:The sum of all possible values of x satisfying the equation is
[2023]
Explanation
Sum of all possible values of x = 3 + (−5/2) = 1/2
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Question for CAT Previous Year Questions: Quadratic equations
Try yourself:Let k be the largest integer such that the equation (x - 1)2 + 2kx + 11 = 0 has no real roots. If y is a positive real number, then the least possible value of k/4y + 9y is
[2023]
Correct Answer : 6
Explanation
Since the equation above has no real roots, the discriminant of the equation should be negative.
The largest integral value that k can take is 4.
Now, we need to minimize k/4y + 9y where k takes the largest integral value and y is positive…
1/y & 9y are both positive real numbers, therefore, their A.M is greater than equal to their G.M.
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Question for CAT Previous Year Questions: Quadratic equations
Try yourself:The price of a precious stone is directly proportional to the square of its weight. Sita has a precious stone weighing 18 units. If she breaks it into four pieces with each piece having distinct integer weight, then the difference between the highest and lowest possible values of the total price of the four pieces will be 288000. Then, the price of the original precious stone is
[2023]
Explanation
“The price of a precious stone is directly proportional to the square of its weight.”
If W is the weight of the stone and P is the price of that stone, then P = k × W2
For the entire, unbroken stone, the price will be 182 × k = 324 k.
“If she breaks it into four pieces with each piece having distinct integer weight, then the difference between the highest and lowest possible values of the total price of the four pieces will be 288000.”
The minimum profit is achieved when the weights of the broken stones are close to each other, that is, the weights are 3, 4, 5, and 6 units.
In this case the combines worth of the four stones = (32 + 42 + 52 + 62) k = 86k
The maximum profit is achieved when the weights of the broken stones are far from each other, that is, the weights are 1, 2, 3, and 12 units.
In this case the combines worth of the four stones = (12 + 22 + 32 + 122) k = 158k
The difference in the total value = 2,88,000.
158k – 86k = 72k = 2,88,000
k = 4,000
So, the price of the original stone = 324 k = 12,96,000
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Question for CAT Previous Year Questions: Quadratic equations
Try yourself:The largest real value of a for which the equation |x + a| + |x - 1| = 2 has an infinite number of solutions for x is
[2022]
Explanation
We are given |x+a| + |x−1| = 2
|x - y| signifies the distance between the points x and y.
|x - (-a)| + |x−1| = 2
Is telling you that the sum of distances between the points x and (-a) and x and 1 is 2.
This implicitly means that the distance between (-a) and 1 can’t be more than 2 in the first place.
Imagine the distance between (-a) and 1 being more than 2.
For this explanation, let us assume that (-a) is to the left of 1.
In this case |x - (-a)| + |x−1| can never be equal to 2 and it will always be greater than 2.
So, the distance between (-a) and 1 can be exactly 2 units or less than 2 units.
Case i: distance between (-a) and 1 is less than 2 units.
Assume the distance between (-a) and 1 is less than 2 units and equal to some ‘x’ units
First of all x can’t be between (-a) and 1, because then |x - (-a)| + |x−1| will be lesser than 2.
x can be y units to the left of (-a) or y units to the right of 1 such that 2y + x = 2 units. But there will just be 2 such solutions and not infinite solutions.
Case ii: distance between (-a) and 1 can be exactly 2 units.
x can only be between (-a) and 1 and can not be anywhere else.
Because if x is not between (-a) or 1, |x - (-a)| + |x−1| will be greater than 2.
Since x can be anywhere between (-a) and 1, the value of |x - (-a)| + |x−1| will be exactly 2 units, this case satisfies the condition!
(-a) can be 2 units to the right of 1 or to the left of 1.
-a = -1
a = 1
(OR)
-a = 3
a = -3 The maximum value of a is 1.
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Question for CAT Previous Year Questions: Quadratic equations
Try yourself:Let a and b be natural numbers. If a2 + ab + a = 14 and b2 + ab + b = 28, then (2a + b) equals
[2022]
Explanation
Since a and b are a + b will also be a natural number, let’s call it k.
k * (k +1) = 42
The product of two consecutive natural numbers is 42.
k = 6.
a + b = 6
a = 2
b = 4.
2a + b = 4 + 4 = 8 2a + b = 8
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Question for CAT Previous Year Questions: Quadratic equations
Try yourself:Let r and c be real numbers. If r and - r are roots of 5x3 + cx2 - 10x + 9 = 0, then c equals
[2022]
Explanation
Since r and (-r) are the roots of the cubic equation 5x3 + cx2 − 10x + 9 = 0
10 r3 - 20 r = 0
r3 - 2r = 0
r (r2 - 2) = 0
Assuming the third root of the cubic equation as k, we can reconstruct it as…
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Question for CAT Previous Year Questions: Quadratic equations
Try yourself:The number of solutions of the equation |x|(6x2 + 1) = 5x2 is
[TITA 2019]
Correct Answer : 5
Explanation
|x| (6x2 + 1) = 5x2
We know that |x2| = x2
Let y = |x|
So, y2 = x2
So, y (6y2 + 1) = 5y2
6y2 + 1 = 5y
6y2 -5 + 1 = 0
y = 1/3 or 1/2
Since, y = |x|
y = 1/3 or -(1/3), (1/2) or -(1/2)
Since we have cancelled y in our first step, x = 0 is also a solution
So, Number of possible solutions = 4 + 1 = 5
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Question for CAT Previous Year Questions: Quadratic equations
Try yourself:The product of the distinct roots of |x2 - x - 6| = x + 2 is
[2019]
Explanation
Given that ∣x2 - x - 6∣ = x + 2
Removing the modulus,
(x2 - x - 6) = x + 2 (or) -(x2 - x - 6) = x + 2
Solving (x2 - x - 6) = x + 2
We get, x2 - 2x - 8 = 0
(x-4) (x+2) = 0
x = 4 or -2
For the solution to be valid, the value of the equation must be positive
When x = 4,
x2 - x - 6 = x + 2
16 - 4 -6 = x + 2
6 - 2 = x
4 = x. Therefore, x = 4 works
For x = -2,
x2 - x - 6 = x +2
4 + 2 - 6 = x + 2
0 - 2 = x
-2 = x. Therefore, x = - 2 works
Similarly, Solving -(x2 - x - 6) = x + 2
-x2 + x + 6 = x + 2
x2 = 4
x = +2 or -2
For the solution to be valid, x2 - x - 6 must be negative
When x = +2,
-x2 + x + 6 = x + 2
-4 + 2 + 6 = x + 2
+ 2 = x Therefore, x = 2 works.
Therefore, Product of distinct roots = 2 x -2 x 4 = -16
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Question for CAT Previous Year Questions: Quadratic equations
Try yourself:What is the largest positive integer such that is also positive integer?
[2019]
Explanation
can be rewritten as
= 1 +
1 + = 1 +
n cannot be -3
has to be an integer
And this value has to be equal to 1, for n to be high
So,= 1
8 = n - 4
n = 12
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Question for CAT Previous Year Questions: Quadratic equations
Try yourself:Let A be a real number. Then the roots of the equation x2 - 4x - log2A = 0 are real and distinct if and only if
[2019]
Explanation
For roots of any Quadratic equation to be real and distinct, D > 0
So, for x2 − 4x - log2A = 0,
D = (-4)2 - (4 x 1 x (-log2A)) > 0
16 + 4 log2A > 0
4 + log2A > 0
log2A > -4
A >
A > (1/16)
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Question for CAT Previous Year Questions: Quadratic equations
Try yourself:The quadratic equation x2 + bx + c = 0 has two roots 4a and 3a, where a is an integer. Which of the following is a possible value of b2 + c?
[2019]
Explanation
Given quadratic equation is x2 + bx + c = 0
Sum of roots = -b
-b = 4a + 3a
-b = 7a
Product of the roots = c
c = 4a x 3a
c = 12a2
b2 = 49a2 and c =12a2
b2 +c = 49a2 + 12a2
b2 +c = 61 a2
Our answer must be a multiple of 61, which should be a perfect square
Going through the options,
549 = 61 x 9
549 = 61 x 32
Thus, possible value of b2 +c = 549
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Question for CAT Previous Year Questions: Quadratic equations
Try yourself:If u2 + (u−2v−1)2 = −4v(u + v), then what is the value of u + 3v?
[2018]
Explanation
Given expression, u2 + (u − 2v − 1)2 = −4v(u + v)
On expanding using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca, we get
u2 + u2 + 4v2 + 1 - 4uv - 2u + 4v = - 4vu - 4v2
2u2 + 4v2 - 2u + 4v + 4v2 + 1 = 0
On rearranging and taking common we get,
2(u2 - u) + 2 (4v2 + 2v) = - 1
Add values on both sides to complete the squares inside parenthesis
2(u2 – u + 1/4) + 2(4v2 + 2v + 1/4) = -1 + 1/2 + 1/2
2(u - 1/2)2 + 2(2v + 1/2)2 = 0
Since they both are squares, they can’t be negative
So, u - 1/2 = 0 and 2v + 1/2 = 0
u = 1/2 and v = −1/4
U+3v =
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Question for CAT Previous Year Questions: Quadratic equations
Try yourself:The minimum possible value of the sum of the squares of the roots of the equation x2 + (a + 3)x - (a + 5) = 0 is
[2017]
Explanation
Let p and q be two roots of the equation
We have to find the minimum value of (p2 + q2)
p2 + q2 = (p + q)2 – 2pq ...(i)
Sum of the roots = (-b)/a.
Therefore, p + q = -(a + 3)
Product of the roots = c/a
Therefore, pq = -(a + 5)
Substituting these values back in (i)
p2 + q2 = [(-a + 3)2 – 2(-a + 5)]
p2 + q2 = a2 + 6a + 9 + 2a + 10
p2 + q2 = a2 + 8a + 19
We have to find the minimum value of a2 + 8a + 19.
By completion of squares, 19 can be split into 16 and 3
⇒ a2 + 8a + 16 + 3
⇒ (a + 4)2 + 3
The minimum possible value is 3 when a = -4
FAQs on Quadratic Equations CAT Previous Year Questions with Answer PDF
1. How do you solve quadratic equations by factoring?
Ans. To solve a quadratic equation by factoring, you need to set the equation equal to zero, factor the quadratic expression, and then set each factor equal to zero to find the roots.
2. Can quadratic equations have imaginary roots?
Ans. Yes, quadratic equations can have imaginary roots if the discriminant (b^2 - 4ac) is negative.
3. What is the quadratic formula used for?
Ans. The quadratic formula is used to find the roots of a quadratic equation ax^2 + bx + c = 0, where a, b, and c are constants.
4. How do you determine the nature of the roots of a quadratic equation?
Ans. The nature of the roots of a quadratic equation can be determined by looking at the discriminant. If the discriminant is positive, the equation has two real roots. If it is zero, the equation has one real root. If it is negative, the equation has two complex roots.
5. Can all quadratic equations be factored?
Ans. Not all quadratic equations can be factored using integers. Some quadratic equations may require the quadratic formula or completing the square to find the roots.