Quadratic formula — Class 10 CBSE Mathematics MCQs with Solutions
Free Class 10 CBSE Mathematics Quadratic formula MCQs with step-by-step solutions (8 questions). Part of Quadratic Equations. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Quadratic formula · easy · theory
By the quadratic formula, the roots of $ax^2 + bx + c = 0$ (where $b^2 - 4ac \geq 0$) are given by:
A. $x = \frac{b \pm \sqrt{b^2 - 4ac}}{2a}$
B. $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ ✓ Correct
C. $x = \frac{-b \pm \sqrt{b^2 + 4ac}}{2a}$
D. $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{a}$
Solution: The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, obtained by completing the square on $ax^2 + bx + c = 0$. Watch the three common slips shown in the wrong options: forgetting the minus sign on $b$, writing $+4ac$ instead of $-4ac$ under the root, and dividing by $a$ instead of $2a$.
Q2 — Quadratic formula · easy · theory
Using the quadratic formula, the roots of $x^2 + 4x + 3 = 0$ are:
A. $1$ and $3$
B. $-1$ and $3$
C. $1$ and $-3$
D. $-1$ and $-3$ ✓ Correct
Solution: Here $a = 1$, $b = 4$, $c = 3$, so $b^2 - 4ac = 16 - 12 = 4$. Then $x = \frac{-4 \pm \sqrt{4}}{2} = \frac{-4 \pm 2}{2}$, giving $x = -1$ and $x = -3$. Check: $(-1)^2 + 4(-1) + 3 = 0$ and $(-3)^2 + 4(-3) + 3 = 0$.
Q3 — Quadratic formula · medium · theory
The roots of $2x^2 - 7x + 3 = 0$, found using the quadratic formula, are:
A. $-3$ and $-\frac{1}{2}$
B. $3$ and $-\frac{1}{2}$
C. $3$ and $\frac{1}{2}$ ✓ Correct
D. $\frac{1}{3}$ and $2$
Solution: Here $a = 2$, $b = -7$, $c = 3$, so the discriminant is $(-7)^2 - 4(2)(3) = 49 - 24 = 25$. Then $x = \frac{7 \pm 5}{4}$, giving $x = 3$ and $x = \frac{1}{2}$. Note that $-b = -(-7) = 7$; keeping the sign wrong would give the negatives of the roots.
Q4 — Quadratic formula · medium · theory
The roots of the equation $x^2 - 4x + 1 = 0$ are:
A. $2 + \sqrt{3}$ and $2 - \sqrt{3}$ ✓ Correct
B. $2 + 2\sqrt{3}$ and $2 - 2\sqrt{3}$
C. $-2 + \sqrt{3}$ and $-2 - \sqrt{3}$
D. $4 + \sqrt{3}$ and $4 - \sqrt{3}$
Solution: The discriminant is $16 - 4 = 12$, so $\sqrt{12} = 2\sqrt{3}$. By the quadratic formula, $x = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}$ — remember to divide both terms of the numerator by 2. These are irrational roots; check: $(2+\sqrt{3})^2 - 4(2+\sqrt{3}) + 1 = 7 + 4\sqrt{3} - 8 - 4\sqrt{3} + 1 = 0$.
Q5 — Quadratic formula · medium · theory
The sum of the squares of two consecutive positive integers is 365. The integers are:
A. $12$ and $13$
B. $13$ and $14$ ✓ Correct
C. $14$ and $15$
D. $15$ and $16$
Solution: Let the integers be $x$ and $x + 1$: $x^2 + (x+1)^2 = 365$ gives $2x^2 + 2x - 364 = 0$, i.e. $x^2 + x - 182 = 0$. By the quadratic formula, $x = \frac{-1 \pm \sqrt{1 + 728}}{2} = \frac{-1 \pm 27}{2}$, so $x = 13$ (rejecting $-14$). The integers are 13 and 14; check: $169 + 196 = 365$.
Q6 — Quadratic formula · hard · theory
A motorboat whose speed in still water is 18 km/h takes 1 hour more to go 24 km upstream than to return downstream to the same spot. The speed of the stream is:
A. $4$ km/h
B. $8$ km/h
C. $9$ km/h
D. $6$ km/h ✓ Correct
Solution: Let the stream's speed be $x$ km/h; upstream speed is $(18 - x)$ and downstream is $(18 + x)$. The condition gives $\frac{24}{18-x} - \frac{24}{18+x} = 1$, which simplifies to $48x = 324 - x^2$, i.e. $x^2 + 48x - 324 = 0$. By the quadratic formula, $x = \frac{-48 \pm \sqrt{2304 + 1296}}{2} = \frac{-48 \pm 60}{2}$, so $x = 6$ km/h (rejecting $-54$).
Q7 — Quadratic formula · hard · theory
The length of a rectangular park is 3 m more than its breadth, and its area is 180 m². The perimeter of the park is:
A. $54$ m ✓ Correct
B. $27$ m
C. $60$ m
D. $42$ m
Solution: Let the breadth be $b$ m; then $b(b + 3) = 180$, i.e. $b^2 + 3b - 180 = 0$. The discriminant is $9 + 720 = 729$ and $\sqrt{729} = 27$, so $b = \frac{-3 + 27}{2} = 12$ m (rejecting the negative root). The length is 15 m, so the perimeter is $2(12 + 15) = 54$ m — note the question asks for the perimeter, not the dimensions.
Q8 — Quadratic formula · hard · theory
Two water taps together can fill a tank in $9\frac{3}{8}$ hours. The tap of larger diameter alone takes 10 hours less than the smaller one to fill the tank. The time taken by the smaller tap alone is:
A. $15$ hours
B. $20$ hours
C. $25$ hours ✓ Correct
D. $12.5$ hours
Solution: Let the smaller tap take $x$ hours; the larger takes $(x - 10)$ hours, and together they fill $\frac{8}{75}$ of the tank per hour since $9\frac{3}{8} = \frac{75}{8}$. So $\frac{1}{x} + \frac{1}{x-10} = \frac{8}{75}$, which gives $8x^2 - 230x + 750 = 0$, i.e. $4x^2 - 115x + 375 = 0$. The quadratic formula gives $x = \frac{115 \pm \sqrt{13225 - 6000}}{8} = \frac{115 \pm 85}{8}$, so $x = 25$ (rejecting $3.75$, which would make $x - 10$ negative). Check: $\frac{1}{25} + \frac{1}{15} = \frac{8}{75}$.