Zeroes & coefficients — Class 10 CBSE Mathematics MCQs with Solutions

Solution: A zero of $p(x)$ is a value of $x$ where $p(x) = 0$, which is exactly a point where the graph of $y = p(x)$ meets the x-axis. Since the graph cuts the x-axis at 3 distinct points, $p(x)$ has exactly 3 zeroes.

Solution: For $ax^2 + bx + c$, the sum of zeroes is $-\frac{b}{a}$. Here $a = 1$ and $b = -7$, so the sum $= -\frac{(-7)}{1} = 7$. Indeed the zeroes are $2$ and $5$, and $2 + 5 = 7$.

Solution: For $ax^2 + bx + c$, the product of zeroes is $\frac{c}{a}$. Here $\frac{c}{a} = \frac{6}{2} = 3$. Note that $4$ is the sum of the zeroes, not the product.

Solution: Sum of zeroes $= 3 + (-4) = -1$ and product $= 3 \times (-4) = -12$. A required polynomial is $x^2 - (\text{sum})x + \text{product} = x^2 - (-1)x + (-12) = x^2 + x - 12$. Check: $(x - 3)(x + 4) = x^2 + x - 12$.

Solution: Setting $x^2 - 3 = 0$ gives $x^2 = 3$, so $x = \pm\sqrt{3}$. We can also write $x^2 - 3 = (x - \sqrt{3})(x + \sqrt{3})$, showing both zeroes clearly.

Solution: Here $\alpha + \beta = 5$ and $\alpha\beta = 6$. Using the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 25 - 12 = 13$. Choosing $25$ forgets to subtract $2\alpha\beta$, while $37$ comes from wrongly adding $2\alpha\beta$.

Solution: Since $x = 2$ is a zero, substituting gives $k(2)^2 + 3(2) + k = 0$, i.e. $4k + 6 + k = 0$. So $5k = -6$, which gives $k = -\frac{6}{5}$.

Solution: From the coefficients, $\alpha + \beta = k + 6$ and $\alpha\beta = 2(2k - 1) = 4k - 2$. The condition $\alpha + \beta = \frac{1}{2}\alpha\beta$ gives $k + 6 = \frac{4k - 2}{2} = 2k - 1$. Solving, $k = 7$.

Solution: A required polynomial is $x^2 - (\text{sum})x + \text{product} = x^2 - \sqrt{2}x + \frac{1}{3}$. Multiplying throughout by $3$ to clear the fraction gives $3x^2 - 3\sqrt{2}x + 1$. Check: sum $= \frac{3\sqrt{2}}{3} = \sqrt{2}$ and product $= \frac{1}{3}$.

Solution: We have $\alpha + \beta = -\frac{5}{2}$ and $\alpha\beta = \frac{1}{2}$. So $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-5/2}{1/2} = -5$.

Solution: Here $\alpha + \beta = 1$ and $\alpha\beta = -2$. Sum of new zeroes $= 2(\alpha + \beta) + 2 = 4$; product $= (2\alpha + 1)(2\beta + 1) = 4\alpha\beta + 2(\alpha + \beta) + 1 = -8 + 2 + 1 = -5$. The required polynomial is $x^2 - 4x - 5$.

Solution: For $ax^3 + bx^2 + cx + d$: $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = \frac{-14}{2} = -7$ and $\alpha\beta\gamma = -\frac{d}{a} = -\frac{8}{2} = -4$. So $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\alpha\beta + \beta\gamma + \gamma\alpha}{\alpha\beta\gamma} = \frac{-7}{-4} = \frac{7}{4}$.

Solution: Sum of zeroes: $(a - d) + a + (a + d) = 3a = -\frac{(-3)}{1} = 3$, so $a = 1$. Product of zeroes: $(a - d) \cdot a \cdot (a + d) = a(a^2 - d^2) = -\frac{1}{1} = -1$, giving $1 - d^2 = -1$, so $d^2 = 2$ and $d = \pm\sqrt{2}$.

Solution: If the zeroes are $\alpha$ and $\frac{1}{\alpha}$, their product is $1$. So $\frac{4k}{k^2 + 4} = 1$, which gives $k^2 - 4k + 4 = 0$, i.e. $(k - 2)^2 = 0$. Hence $k = 2$ only; $k = -2$ gives product $\frac{-8}{8} = -1$, which does not work.

Solution: From the polynomial, $\alpha + \beta = 6$. Writing $3\alpha + 2\beta = 2(\alpha + \beta) + \alpha = 12 + \alpha = 20$ gives $\alpha = 8$, so $\beta = -2$. Then $k = \alpha\beta = 8 \times (-2) = -16$.