Test: Quadratic Equations (Easy) - Class 10 MCQ

x² - 3x = 10

x² - 5x + 2x - 10 = 0

(x - 5)(x + 2) = 0

Therefore x = 5

The maximum exponent of variable x is 2. So it is a quadratic equation.

Alternate (B) : 3x² - 4x + 1 = 0

The maximum exponent of variable x is 2. So it is a quadratic equation.

Alternate (C) : 2x - 7 = 0

The maximum exponent of variable x is 1. So it is not a quadratic equation.

Given,

Hence, 6x - 5x² = 7

⇒ 5x² - 6x + 7 = 0

⇒ ax² + bx + c = 0

Hence,it is a quadratic equation.

b² – 4ac = 4² – 4 x 5 x 5

= 16 – 100

= -84 < 0

⟹ x² + 1 + 2x – x² = 0

⟹ 1 + 2x = 0

⟹ x = -1 / 2

This gives only 1 real value of x.

Then according question, x + 12 = 160 / x

x² + 12x – 160 = 0

x² + 20x – 8x – 160 = 0

(x + 20) (x – 8) = 0

x = -20, 8

Since the number is natural, so we consider only positive value.

According to the question:

5n × 5(n + 1) = 300

⇒ n² + n – 12 = 0

⇒ (n – 3) (n + 4) = 0

⇒ n = 3 and n = – 4.

As n is a positive natural number so n = – 4 will be discarded.

Therefore the numbers are 15 and 20.

⇒ p²x² = q²

⇒ x = ±p / q

4 = p - 4

P = 8

Therefore,

9 (x + 10) = x²

⇒ x² – 9x – 90 = 0

⇒ x² – 15x + 6x – 90 = 0

⇒ x(x – 15) + 6 (x – 15) = 0

⇒ (x + 6) (x – 15) = 0

Therefore x = – 6 or x =15

Since x is the marks obtained, x ≠ – 6.

Therefore, x = 15.

X² + 2x + 1 - x² = 0

2x + 1 = 0

x = -½

Hence, there is one real root.

100x² – 10x – 10x + 1 = 0

10x(10x – 1) -1(10x – 1) = 0

(10x – 1)² = 0

∴ (10x – 1) = 0 or (10x – 1) = 0

⇒ x = 1 / 10 or x = 1 / 10