Class 10 The Quadratic Formula — practice questions with answers

Quadratic Equations · 5 sample questions from the school syllabus. Try each one, then open the answer and the worked solution.

1. The quadratic formula gives the roots of ax² + bx + c = 0 (a not 0). Which is the correct formula?

A $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$
B $x = \frac{b \pm b ^{2} - 4 a c}{2 a}$
C $x = \frac{- b \pm b ^{2} + 4 a c}{2 a}$
D $x = \frac{- b \pm b ^{2} - 4 a c}{a}$

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Answer: A)

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

Completing the square on ax² + bx + c = 0 gives

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

2. If b² - 4ac < 0 for ax² + bx + c = 0, then the equation has:

Ano real roots
Btwo equal roots
Ctwo distinct real roots
Dexactly one real root

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Answer: A) no real roots

A negative discriminant means the square root is not real, so there are no real roots.

3. If one root of x² - 8x + k = 0 is 3, find the other root using the formula and the value of k.

Aother root 5, k = 15
Bother root 5, k = 24
Cother root 11, k = 15
Dother root 2, k = 6

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Answer: A) other root 5, k = 15

If 3 is a root, 9 - 24 + k = 0 => k = 15. Then x² - 8x + 15 = 0 =>

x = \frac{8 \pm 64 - 60}{2} = \frac{8 \pm 2}{2}

= 5 or 3. Other root 5.

4. An aeroplane covers 1500 km at a steady speed. Bad weather forces the pilot to fly 100 km/h slower than usual, and the flight then takes 30 minutes longer.
Taking the usual speed as x km/h, the situation reduces to x² − 100x − 300000 = 0.
Find the usual speed.

A500 km/h
B700 km/h
C600 km/h
D−500 km/h

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Answer: C) 600 km/h

\frac{1500}{x - 100} - \frac{1500}{x} = \frac{1}{2}

clears to x² − 100x − 300000 = 0.

x = \frac{100 \pm 10000 + 1200000}{2} = \frac{100 \pm 1100}{2}

= 600 or −500. The root −500 satisfies the equation but a speed cannot be negative, so it is rejected: the usual speed is 600 km/h. Check: 1500/500 − 1500/600 = 3 − 2.5 = 0.5 hour.

5. Solve 3x² + 8x − 3 = 0 with the formula, then check by putting each root back into the equation. The roots are:

A−3 and $- \frac{1}{3}$
B3 and $\frac{1}{3}$
C3 and $- \frac{1}{3}$
D $\frac{1}{3}$ and −3

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Answer: D)

\frac{1}{3}

and −3

x = \frac{- 8 \pm 64 + 36}{6} = \frac{- 8 \pm 10}{6}

\frac{2}{6} = \frac{1}{3}

\frac{- 18}{6} = - 3

. Substituting back:

3 \times \frac{1}{9} + \frac{8}{3} - 3 = \frac{1}{3} + \frac{8}{3} - 3 = 0

, and 3(9) − 24 − 3 = 0. Substituting is what catches a swap such as 3 and

- \frac{1}{3}

, since 27 + 24 − 3 is nowhere near 0.

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