Zeroes & coefficients — Class 10 CBSE Mathematics MCQs with Solutions

Solution: By definition, a zero of a polynomial $p(x)$ is a value $a$ for which $p(a) = 0$. This is the fundamental concept: when you substitute the zero into the polynomial, the output is exactly zero. The other options confuse zeroes with other polynomial properties (domain, constant term, degree).

Solution: When a polynomial's graph crosses the x-axis at a point, the x-coordinate of that point is a real zero. Two distinct crossing points mean two distinct real zeroes. Touching the axis (not crossing) would indicate a repeated zero; complex zeroes don't appear as x-intercepts on a real graph.

Solution: The relationship between zeroes and coefficients is derived from the factored form: $p(x) = a(x - \alpha)(x - \beta)$. Expanding gives $p(x) = ax^2 - a(\alpha + \beta)x + a\alpha\beta$. Comparing with $ax^2 + bx + c$, we get $b = -a(\alpha + \beta)$, so $\alpha + \beta = -\frac{b}{a}$. This is a KEY relationship to memorize and understand.

Solution: A cubic polynomial always has degree 3 and always has at least one real zero (because odd-degree polynomials have end behaviours that force them to cross the x-axis). By the Fundamental Theorem of Algebra, it has exactly 3 zeroes counting multiplicity (in complex numbers). However, these three zeroes need not be DISTINCT — one could be repeated, or some could be non-real complex conjugates. So a cubic need not have three distinct real zeroes.

Solution: A fundamental theorem in algebra states that a polynomial of degree $n$ has at most $n$ zeroes (counting multiplicity). This is because if $p(x)$ has $n$ distinct zeroes $r_1, \ldots, r_n$, then $p(x) = a(x - r_1) \cdots (x - r_n)$, which has degree exactly $n$. More zeroes would require a higher-degree polynomial.

Solution: For $p(x) = x^2 - 5x + 6$, we have $a = 1$, $b = -5$, $c = 6$. Using Vieta's formulas: $\alpha + \beta = -\frac{b}{a} = -\frac{(-5)}{1} = 5$ and $\alpha \beta = \frac{c}{a} = \frac{6}{1} = 6$. Note: you could verify by factoring ($p(x) = (x - 2)(x - 3)$, so zeroes are 2 and 3, and indeed $2 + 3 = 5$ and $2 \times 3 = 6$), but the key is understanding the relationship from the formula.

Solution: Using Vieta's formula $\alpha + \beta = -\frac{b}{a}$ with $a = 1$, we have $7 = -b$, so $b = -7$. We can determine $b$ uniquely from the sum of zeroes, but we cannot determine $c$ (the product of zeroes) from the sum alone — that would require additional information. The relationship between zeroes and the coefficient of $x$ is always fixed.

Solution: If $\alpha$ and $\beta$ are zeroes, the polynomial can be written as $p(x) = a(x - \alpha)(x - \beta) = a[x^2 - (\alpha + \beta)x + \alpha\beta]$. With $a = 1$, $\alpha + \beta = 4$, and $\alpha\beta = 3$, we get $p(x) = x^2 - 4x + 3$. This demonstrates the reverse problem: given zeroes (or their sum and product), construct the polynomial. The key is understanding how Vieta's formulas work in both directions.

Solution: Using the algebraic identity: $(\alpha + \beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2$, so $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$. This shows how to express the sum of squares of zeroes using Vieta's formulas. Since $\alpha + \beta = -b$ and $\alpha\beta = c$, we have $\alpha^2 + \beta^2 = (-b)^2 - 2c = b^2 - 2c$. This type of relationship (involving symmetric functions of zeroes) is fundamental in understanding polynomial structure.

Solution: For $x^2 - 3x + 2$: $\alpha + \beta = 3$ and $\alpha\beta = 2$. So $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{3}{2}$. This demonstrates finding the sum of reciprocals of zeroes using Vieta's formulas. The key algebraic step is: $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta + \alpha}{\alpha\beta}$. This is a standard technique for converting relationships between zeroes into expressions in terms of coefficients.

Solution: We have $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta}$. Using Vieta's formulas: $\alpha + \beta = -p$ and $\alpha\beta = q$. We know $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = p^2 - 2q$. So $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{p^2 - 2q}{q^2}$. This shows how symmetric expressions in the zeroes can always be reduced to expressions in the coefficients — a powerful technique in polynomial theory.

Solution: By Vieta's formulas, the product of the zeroes equals the constant term divided by the leading coefficient: $\alpha \beta = \frac{q}{1} = q$. If the product is zero, then $q = 0$. This means the polynomial is $x^2 + px = x(x + p)$, which has zeroes $0$ and $-p$. Notice: when the constant term is zero, one zero is always $0$; when the coefficient of $x$ is zero, the non-zero zeroes are opposites.

Solution: Set $p(x) = 0$ and factorize by splitting the middle term: $x^2 - 5x + 2x - 10 = 0$. Factor: $(x - 5)(x + 2) = 0$. Therefore $x = 5$ or $x = -2$.

Solution: Let the zeros be $\alpha$ and $\frac{1}{\alpha}$. Product of zeros = $\frac{k}{3}$. Since $\alpha \cdot \frac{1}{\alpha} = 1$, we have $1 = \frac{k}{3}$, so $k = 3$.

Solution: Sum of zeros: $S = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4$. Product of zeros: $P = (2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1$. Polynomial: $x^2 - 4x + 1$.

Solution: From $p(x) = x^2 - 5x + 6$: $\alpha + \beta = 5$ and $\alpha\beta = 6$. Evaluate: $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{5}{6}$.

Solution: When a parabola touches the $x$-axis at exactly one point, the discriminant $b^2 - 4ac = 0$, which means the polynomial has two equal (coincident) real zeros.

Solution: Sum of zeros: $-\frac{2}{k}$. Product of zeros: $\frac{3k}{k} = 3$. Given sum = product: $-\frac{2}{k} = 3$. Solve: $3k = -2$, so $k = -\frac{2}{3}$.

Solution: From $4x^2 - 3x - 7$: $\alpha + \beta = \frac{3}{4}$ and $\alpha\beta = -\frac{7}{4}$. Calculate: $\alpha + \beta + \alpha\beta = \frac{3}{4} - \frac{7}{4} = -1$.

Solution: The general form of a quadratic polynomial with given zeros $\alpha$ and $\beta$ is $p(x) = k[x^2 - (\alpha + \beta)x + \alpha\beta]$ where $k$ is any non-zero real constant. Since $k$ can take infinitely many values, there are infinitely many polynomials.

Solution: For $p(x) = x^2 - 4x + k$, the product of zeros equals $k$. Given that the product is $3$, therefore $k = 3$.

Solution: Using formula $x^2 - Sx + P$ where $S$ is sum and $P$ is product: $x^2 - (-3)x + 2 = x^2 + 3x + 2$.

Solution: From $2x^2 + 7x + 5$: $\alpha + \beta = -\frac{7}{2}$ and $\alpha\beta = \frac{5}{2}$. Using $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \frac{49}{4} - 5 = \frac{29}{4}$.

Solution: Sum of zeros: $2 + (-3) = -1 = -(a+1)$, so $a + 1 = 1$ and $a = 0$. Product of zeros: $2 \times (-3) = -6 = b$.

Solution: Let zeros be $\alpha$ and $-\alpha$. Their sum is $\alpha + (-\alpha) = 0$. Using sum formula: $-\frac{b}{a} = 0$, therefore $b = 0$.

Solution: From $x^2 - 4x + 1$: $\alpha + \beta = 4$ and $\alpha\beta = 1$. Using $\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = 64 - 3(1)(4) = 64 - 12 = 52$.

Solution: Sum of zeros: $\alpha + \beta = -99$ (negative). Product of zeros: $\alpha\beta = 127$ (positive). Since the product is positive, both zeros have the same sign. Since their sum is negative, both zeros must be negative.

Solution: By the Fundamental Theorem of Algebra, a polynomial of degree $n$ has at most $n$ real zeros (roots).

Solution: From $3x^2 - 6x + 4$: $\alpha + \beta = 2$ and $\alpha\beta = \frac{4}{3}$. Calculate: $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{4 - \frac{8}{3}}{\frac{4}{3}} = \frac{\frac{4}{3}}{\frac{4}{3}} = 1$.

Solution: From $x^2 + x - 2$: $\alpha + \beta = -1$ and $\alpha\beta = -2$. Calculate: $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = 1 + 8 = 9$, so $\alpha - \beta = \pm 3$. Therefore $\frac{1}{\alpha} - \frac{1}{\beta} = \frac{\beta - \alpha}{\alpha\beta} = \frac{\mp 3}{-2} = \pm \frac{3}{2}$.