Mathematics — Standard (041) — CBSE Class 10 board question

We need to find the zeroes of $p(x) = x^2 + \dfrac{1}{2}x - 1$.

Multiply throughout by 2: $2x^2 + x - 2$...

Instead, split the middle term directly:

$$x^2 + \frac{1}{2}x - 1 = x^2 + x - \frac{1}{2}x - 1 = x(x+1) - \frac{1}{2}(x+1) = \left(x - \frac{1}{2}\right)(x+1)$$

So $p(x) = 0$ when $x = \dfrac{1}{2}$ or $x = -1$.

The zeroes of $p(x)$ are $\dfrac{1}{2}$ and $-1$.

Source: Chapter 2, Section 2.3

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• The key step is splitting the middle term: $\frac{1}{2}x = x - \frac{1}{2}x$, so that the product of the split terms equals $1 \times (-1) = -1$ (coefficient of $x^2 \times$ constant term).
• Examiners expect you to show the factorisation step clearly before stating the zeroes.
• For a 2-mark question: 1 mark for correct factorisation, 1 mark for stating both zeroes.
• You can verify: sum of zeroes $= \frac{1}{2} + (-1) = -\frac{1}{2} = \frac{-b}{a}$ ✓ and product $= \frac{1}{2} \times (-1) = -1 = \frac{c}{a}$ ✓ (though verification is not asked here).