Test: Quadratic Equations (Hard) - Class 10 MCQ

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.

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

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

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.

⇒ (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

x(x + 5) - 3(x + 5) = 0

(x - 3)(x + 5) = 0

=> x = 3 or x = -5.

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

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.

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

Sum of roots = -(-12) = 12

a + 3a = 4a = 12

=> a = 3

Product of the roots = 3a² = 3(3)² = 27.

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.

Step 2: The roots of quadratic equation are real only when D ≥ 0 16 – 4c ≥ 0

Step 3: c ≤ 4

D = 1 > 0 ⇒ Two distinct real roots.

= (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.

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

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