Mathematics — CBSE Class 10 board question

Finding zeroes:

Using the identity $a^2 - b^2 = (a-b)(a+b)$:

$$x^2 - 15 = \left(x - \sqrt{15}\right)\left(x + \sqrt{15}\right)$$

So, $x^2 - 15 = 0$ when $x = \sqrt{15}$ or $x = -\sqrt{15}$.

The zeroes are $\alpha = \sqrt{15}$ and $\beta = -\sqrt{15}$.

Verification:

Here $a = 1$, $b = 0$, $c = -15$.

$$\text{Sum of zeroes} = \sqrt{15} + (-\sqrt{15}) = 0 = \frac{-0}{1} = \frac{-b}{a} \checkmark$$

$$\text{Product of zeroes} = \sqrt{15} \times (-\sqrt{15}) = -15 = \frac{-15}{1} = \frac{c}{a} \checkmark$$

Hence, the relationship is verified.

Source: Chapter 2, Section 2.3

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• This is modelled on Example 3 of the textbook (which uses $x^2 - 3$); the same method applies to $x^2 - 15$.
• Examiners award marks for: (1) correct factorisation/zeroes, (2) correct sum verification, (3) correct product verification. All three steps must be shown explicitly.
• Write $\frac{-b}{a}$ and $\frac{c}{a}$ with actual coefficient values substituted — don't just state the formula.
• The coefficient of $x$ is 0 (not absent), so $b = 0$; state this clearly to avoid losing marks.