Standard form — Class 10 CBSE Mathematics MCQs with Solutions
Free Class 10 CBSE Mathematics Standard form 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 — Standard form · easy · theory
Which of the following is a quadratic equation?
A. $2x + 3 = 0$
B. $x^2 - 5x + 6 = 0$ ✓ Correct
C. $x^3 + 2x^2 + x = 0$
D. $\frac{1}{x} + 2 = 0$
Solution: A quadratic equation has the form $ax^2 + bx + c = 0$ where $a \neq 0$ and the highest power of $x$ is 2. In $x^2 - 5x + 6 = 0$, we have $a=1$, $b=-5$, $c=6$, making it quadratic. The other equations are linear, cubic, or rational—not quadratic.
Q2 — Standard form · easy · theory
Why is the condition $a \neq 0$ essential in the standard form $ax^2 + bx + c = 0$ of a quadratic equation?
A. Because we need to divide by $a$ in the quadratic formula ✓ Correct
B. Because the equation would be too easy to solve
C. Because $a$ must always be a positive integer
D. To make sure there are always two real roots
Solution: If $a = 0$, the equation $0 \cdot x^2 + bx + c = 0$ becomes $bx + c = 0$, which is linear, not quadratic. Additionally, the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ requires $a \neq 0$ in the denominator. The condition $a \neq 0$ is what distinguishes a quadratic equation from a linear one.
Q3 — Standard form · easy · theory
For a quadratic equation $ax^2 + bx + c = 0$, if the roots are $\alpha$ and $\beta$, which expression gives the sum of the roots?
A. $\frac{c}{a}$
B. $-\frac{b}{a}$ ✓ Correct
C. $\frac{b}{c}$
D. $\frac{a}{b}$
Solution: By Vieta's formulas, for roots $\alpha$ and $\beta$ of $ax^2 + bx + c = 0$, we have $\alpha + \beta = -\frac{b}{a}$. This relationship connects the sum of roots directly to the coefficients, without needing to compute the roots themselves.
Q4 — Standard form · easy · theory
For the quadratic equation $2x^2 - 8x + 6 = 0$, what is the product of the roots?
A. $-\frac{8}{2} = -4$
B. $\frac{6}{2} = 3$ ✓ Correct
C. $\frac{-8}{6}$
D. $\frac{2}{6}$
Solution: By Vieta's formulas, the product of roots $\alpha \cdot \beta = \frac{c}{a}$. In $2x^2 - 8x + 6 = 0$, we have $a=2$ and $c=6$, so the product is $\frac{6}{2} = 3$. Note: the sum of roots would be $-\frac{-8}{2} = 4$, not the product.
Q5 — Standard form · medium · theory
If a quadratic has roots $\alpha = 2$ and $\beta = -3$, construct the quadratic equation using the relationship $\alpha + \beta$ and $\alpha \beta$.
A. $x^2 + x - 6 = 0$ ✓ Correct
B. $x^2 - x - 6 = 0$
C. $x^2 - 5x + 6 = 0$
D. $x^2 + 5x - 6 = 0$
Solution: The sum of roots is $\alpha + \beta = 2 + (-3) = -1$, and the product is $\alpha \beta = 2 \cdot (-3) = -6$. By Vieta's formulas (reversed), the equation is $x^2 - (\text{sum})x + (\text{product}) = 0$, giving $x^2 - (-1)x + (-6) = 0$ or $x^2 + x - 6 = 0$. This demonstrates reconstructing a quadratic from root properties.
Q6 — Standard form · hard · theory
Derive the relationship $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$ for the roots of a quadratic, and explain why this is useful.
A. It allows us to express the sum of squares of roots entirely in terms of the coefficients, without finding the roots themselves ✓ Correct
B. It proves that the roots must always be real
C. It shows that the discriminant is always positive
D. It simplifies the quadratic formula
Solution: Expanding $(\alpha + \beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2$, we get $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$. Using Vieta's formulas: $\alpha^2 + \beta^2 = (-\frac{b}{a})^2 - 2(\frac{c}{a}) = \frac{b^2}{a^2} - \frac{2c}{a} = \frac{b^2-2ac}{a^2}$. This expresses a property of roots using only coefficients—a powerful technique in algebra.
Q7 — Standard form · hard · theory
A quadratic has roots whose sum is 7 and product is 10. Explain why these roots must be real and distinct, and write the quadratic.
A. Because the discriminant $= (\text{sum})^2 - 4(\text{product}) = 49 - 40 = 9 > 0$. The quadratic is $x^2 - 7x + 10 = 0$. ✓ Correct
B. Because the roots are always real when they are integers; the quadratic is $x^2 + 7x + 10 = 0$.
C. Because the sum exceeds the product; the quadratic is $x^2 - 7x + 10 = 0$.
D. There is no way to determine if roots are real from this information; the quadratic is $x^2 + 7x - 10 = 0$.
Solution: For a monic quadratic $x^2 + px + q = 0$ with roots whose sum is $S$ and product is $P$, we have $p = -S$ and $q = P$. Here, $p = -7$ and $q = 10$, giving $x^2 - 7x + 10 = 0$. The discriminant is $(-7)^2 - 4(1)(10) = 49 - 40 = 9 > 0$, confirming two distinct real roots.
Q8 — Standard form · hard · theory
Analyse the following claim: 'If the coefficients $a$, $b$, $c$ of $ax^2 + bx + c = 0$ are all positive, the quadratic has no positive real roots.' Is this true, and why?
A. Yes, because for any positive $x$: $ax^2 + bx + c = a(x^2) + b(x) + c > 0$, so $ax^2 + bx + c \neq 0$ for positive $x$. ✓ Correct
B. No, because positive coefficients always produce positive roots.
C. Yes, but only if $a > 1$.
D. No, because the sign of coefficients does not determine the sign of roots.
Solution: If $a > 0$, $b > 0$, $c > 0$, and $x > 0$, then $ax^2 > 0$, $bx > 0$, and $c > 0$, so $ax^2 + bx + c > 0$. The quadratic is always positive for positive $x$, meaning the parabola never crosses the x-axis on the right side. Any real roots must be non-positive. This illustrates Descartes' Rule of Signs: the number of positive roots equals the number of sign changes in the coefficient sequence.