Standard form — Class 10 CBSE Mathematics MCQs with Solutions
Free Class 10 CBSE Mathematics Standard form MCQs with step-by-step solutions (6 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. $x^2 + 3\sqrt{x} + 2 = 0$
B. $(x+1)^2 = 2(x-3)$ ✓ Correct
C. $(x-2)(x+1) = (x-1)(x+3)$
D. $x^2 + \frac{1}{x^2} = 5$
Solution: Expanding $(x+1)^2 = 2(x-3)$ gives $x^2 + 2x + 1 = 2x - 6$, i.e. $x^2 + 7 = 0$, which is a polynomial equation of degree 2, so it is quadratic. The first option contains $\sqrt{x}$, so it is not a polynomial equation. The third simplifies to $-3x + 1 = 0$ (linear), and the fourth becomes $x^4 - 5x^2 + 1 = 0$ (degree 4).
Q2 — Standard form · easy · theory
A quadratic equation in the variable $x$ is of the form $ax^2 + bx + c = 0$, where $a$, $b$, $c$ are real numbers and:
A. $a \neq 0$ ✓ Correct
B. $b \neq 0$
C. $c \neq 0$
D. $a > 0$
Solution: If $a = 0$, the term $ax^2$ vanishes and the equation becomes $bx + c = 0$, which is linear, not quadratic. So the only condition needed is $a \neq 0$. The values of $b$ and $c$ may be zero, and $a$ may be negative, e.g. $-2x^2 + 5 = 0$ is a valid quadratic equation.
Q3 — Standard form · easy · theory
Which of the following values of $x$ is a root of the equation $x^2 - 5x + 6 = 0$?
A. $x = 1$
B. $x = 4$
C. $x = 2$ ✓ Correct
D. $x = 5$
Solution: A value is a root if it satisfies the equation. Substituting $x = 2$: $(2)^2 - 5(2) + 6 = 4 - 10 + 6 = 0$, so $x = 2$ is a root. Checking the others: $x = 1$ gives $2$, $x = 4$ gives $2$, and $x = 5$ gives $6$ — none of them is zero.
Q4 — Standard form · medium · theory
If $x = \frac{1}{2}$ is a root of the equation $2x^2 + kx - 3 = 0$, then the value of $k$ is:
A. $-5$
B. $2$
C. $-2$
D. $5$ ✓ Correct
Solution: Since $x = \frac{1}{2}$ is a root, it satisfies the equation: $2\left(\frac{1}{4}\right) + \frac{k}{2} - 3 = 0$. This gives $\frac{1}{2} + \frac{k}{2} = 3$, so $\frac{k}{2} = \frac{5}{2}$ and $k = 5$. Check: $2x^2 + 5x - 3 = (2x - 1)(x + 3)$, which is indeed zero at $x = \frac{1}{2}$.
Q5 — Standard form · medium · theory
When the equation $(x-3)(2x+1) = x(x+5)$ is written in the standard form $ax^2 + bx + c = 0$, it becomes:
A. $x^2 - 10x - 3 = 0$ ✓ Correct
B. $x^2 + 10x - 3 = 0$
C. $3x^2 - 10x - 3 = 0$
D. $x^2 - 10x + 3 = 0$
Solution: Expanding the left side: $(x-3)(2x+1) = 2x^2 - 5x - 3$. The right side is $x^2 + 5x$. Bringing everything to one side: $2x^2 - 5x - 3 - x^2 - 5x = 0$, which simplifies to $x^2 - 10x - 3 = 0$. Since the degree is 2, the given equation is quadratic.
Q6 — Standard form · hard · theory
A train travels 480 km at a uniform speed. If its speed had been 8 km/h less, it would have taken 3 hours more to cover the same distance. Taking the usual speed as $x$ km/h, the situation is represented by the quadratic equation:
A. $x^2 + 8x - 1280 = 0$
B. $x^2 - 8x + 1280 = 0$
C. $x^2 - 8x - 1280 = 0$ ✓ Correct
D. $3x^2 - 8x - 480 = 0$
Solution: Time at usual speed is $\frac{480}{x}$ hours and at the reduced speed it is $\frac{480}{x-8}$ hours, and the slower journey takes 3 hours more: $\frac{480}{x-8} - \frac{480}{x} = 3$. Multiplying through by $x(x-8)$: $480x - 480(x-8) = 3x(x-8)$, i.e. $3840 = 3x^2 - 24x$. Dividing by 3 gives $x^2 - 8x - 1280 = 0$.