Test: Quadratic Equations (Hard)
20 Questions MCQ Test Additional Documents & Tests for Class 10 | Test: Quadratic Equations (Hard)
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The values of k for which the roots are real and equal of the following equation kx2 —
2√5x + 4 = 0 is k = 5/4
Which of the following is not quadratic equation?
x2 – 4x + 4 + 1 = 2x – 3
x2 – 4x + 4 + 1 – 2x + 3 = 0
x2 – 6x + 8 = 0
This is a quadratic equation.
(b) (x + 2)3 = x3 – 4
x3 + 6x2 + 12x + 8 = x3 – 4
6x2 + 12x + 12 = 0
This is a quadratic equation.
(c) x(2x + 3) = x2 + 1
2x2 + 3x = x2 + 1
x2 + 3x - 1 = 0
This is a quadratic equation.
(d) x(x + 1) + 8 = (x + 2) (x – 2)
x2 + x + 8 = x2 – 4
x + 12 = 0
This is not a Quadratic equation.
If the sum of the roots of a quadratic equation is 5 and the product of the roots is also 5, then the equation is
sum of roots = −ba
product of roots = ca.
sum of roots = 5 = −ba
product of roots = 5 = ca,
Thus, quadratic equation is x2 − 5x + 5 = 0
A rectangular field has an area of 3 sq. units. The length is one more than twice the breadth ‘x’. Frame an equation to represent this.
Given, length = (2 × breadth + 1)
Let the breadth of the field be x.
Length of the field = 2x + 1
Area of the rectangular field = x (2x + 1) =3
2x2 + x = 3
2x2 + x − 3 = 0
If a train travelled 5 km/hr faster, it would take one hour less to travel 210 km. The speed of the train is :
Distance travelled = 210 km
Time taken to travel 210 km = 210 / x hours
When the speed is increased by 5 km/h, the new speed is (x + 5)
Time taken to travel 210 km with the new speed is 210 / (x + 5) hours
According to the question,
210 / x − 210 / (x + 5) = 1
⇒ 210(x + 5) − 210x = x(x + 5)
⇒ 210x + 1050 − 210x = x2 + 5x
⇒ x2 + 5x − 1050 = 0
⇒ (x + 35)(x − 30) = 0
⇒ x = −35, 30
The speed cannot be negative.
Thus, the speed of the train is 30 km/hr.
If the solutions of the equation x2 + 3x − 18 = 0 are -6, 3 then the roots of the equation 2(x2 + 3x − 18) = 0 are
⇒ (5x + 1)2 = 16 (Applying (a + b)2 formula)
⇒ 5x + 1 = ± 4(Taking square root on both sides)
⇒ 5x = -5, 3
⇒ x = −1, 3 / 5
Find the roots of the quadratic equation: x2 + 2x - 15 = 0?
x(x + 5) - 3(x + 5) = 0
(x - 3)(x + 5) = 0
=> x = 3 or x = -5.
Using the method of completion of squares find one of the roots of the equation 2x2 − 7x + 3 = 0. Also, find the equation obtained after completion of the square.
Dividing by the coefficient of x2, we get
x2 − 7 / 2x + 3 / 2 = 0; a = 1, b = 7 / 2, c = 3 / 2
Adding and subtracting the square of b / 2 = 7 / 4, (half of coefficient of x)
We get, [x2 − 2 (7 / 4) x + (7 / 4)2] − (7 / 4)2 + 3 / 2 = 0
The equation after completing the square is: (x − 7 / 4)2 − 25 / 16 = 0
Taking square root, (x − 7 / 4) = (±5 / 4)
Taking positive sign 5 / 4, x = 3
Taking negative sign −5 / 4, x = 1 / 2
If the roots of a quadratic equation are 20 and -7, then find the equation?
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 roots of the equation 5x2–6x–2=0 by the method of completing the square.
This is the same as:
(5x)2 – [2× (5x) ×3] +32 – 32 – 10 = 0
⇒ (5x – 3)2 – 9 – 10 = 0
⇒ (5x – 3)2 – 19 = 0
⇒ (5x – 3)2 = 19
⇒ 5x − 3 = ± √19
⇒ x = (3± √19) / 5
One root of the quadratic equation x2 - 12x + a = 0, is thrice the other. Find the value of a?
Sum of roots = -(-12) = 12
a + 3a = 4a = 12
=> a = 3
Product of the roots = 3a2 = 3(3)2 = 27.
There is a natural number x. Write down the expression for the product of x and its next natural number.
What number should be added to x2 + 6x to make it a perfect square?
If we observe carefully we can see that x2 + 6x can be written in the form of
(a2 + 2ab + b2) by adding a constant. x2 + 2(x) (3) +constant.
To make x2 + 6x a perfect square, divide the coefficient of x by 2 and then add the square of the result to make this a perfect square.
Hence, 6 / 2 = 3 and 32 = 9
We should add 9 to make x2 + 6x a perfect square.
The equation x2 + 4x + c = 0 has real roots, then
Step 2: The roots of quadratic equation are real only when D ≥ 0 16 – 4c ≥ 0
Step 3: c ≤ 4
Find the discriminant of the quadratic equation 3x2 – 5x + 2 = 0 and hence, find the nature of the roots.
D = 1 > 0 ⇒ Two distinct real roots.
Find the value of a/b + b/a, if a and b are the roots of the quadratic equation x2 + 8x + 4 = 0?
= (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.
If the equation x2+2(k+2) x+9k=0 has equal roots, then values of k are __________.
Step 2: The roots of quadratic equation are real and equal only when D = 0
k2 + 4 − 5k = 0
⇒ k2 − 5k + 4 = 0
⇒ k2 − k − 4k + 4 = 0
⇒ k(k − 1) − 4(k − 1) = 0
⇒ (k − 1)(k − 4) = 0
Step 3: k = 4 or 1
Find the roots of the 3x2 – 5x + 2 = 0 quadratic equation, using the quadratic formula.
Quadratic equation of the form ax2 + bx + c = 0
The roots of the above quadratic equation will be
a = 3, b = -5 and c = 2
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