Vardaan Learning Institute

Ans: (b) 1. A straight line (linear polynomial) meets the $x$-axis at exactly one point, so it has exactly 1 zero.

Ans: (b) 1. If the parabola only touches (does not cross) the $x$-axis, the two zeroes coincide — it has 1 repeated zero.

Ans: (a) 2 and 3. $x^2-5x+6 = (x-2)(x-3)$. Zeroes: $x=2$ and $x=3$.

Ans: (b) 4. $\alpha+\beta = -\dfrac{b}{a} = -\dfrac{-4}{1} = 4$.

Ans: (b) $\dfrac{3}{2}$. $\alpha\beta = \dfrac{c}{a} = \dfrac{3}{2}$.

Ans: (c) $x^2 - 3x + 2$. $p(x) = x^2 - (\text{sum})x + (\text{product}) = x^2 - 3x + 2$.

Ans: (b) 5. If zeroes are $\alpha$ and $\dfrac{1}{\alpha}$, product $= 1 = \dfrac{k}{5} \implies k = 5$.

Ans: (c) No real zeroes. Zeroes of a polynomial are the $x$-coordinates of the points where $y = p(x)$ meets the $x$-axis. If it never does, there are no real zeroes.

Ans: (c) 1 and –1. $x^2-1 = (x-1)(x+1)$. Zeroes: $1$ and $-1$.

Ans: (d) $-\dfrac{2}{3}$. Sum $= -\dfrac{2}{k}$; Product $= \dfrac{3k}{k} = 3$. Given sum = product: $-\dfrac{2}{k} = 3 \implies k = -\dfrac{2}{3}$.

Ans: (b) 2. The highest power of $x$ in $\sqrt{3}\,x^2 - 8$ is 2, so degree = 2.

Ans: (a) 1. $p(2) = 4 + 2k - 6 = 0 \implies 2k - 2 = 0 \implies k = 1$.

Ans:

1. Touches $x$-axis at one point (does not cross): 1 zero (a repeated root — the vertex of the parabola lies on the $x$-axis).
2. Cuts $x$-axis at two distinct points: 2 zeroes.
3. Entirely below $x$-axis, never touches: No real zeroes.
4. Cuts $x$-axis at three points: 3 zeroes (this is a cubic polynomial).

Ans:

1. The graph crosses the $x$-axis at 2 points, so $p(x)$ has 2 zeroes.
2. Zeroes of $p(x)$ are $\mathbf{-2}$ and $\mathbf{3}$.
3. Sum of zeroes $= -2+3 = 1$; Product $= -2 \times 3 = -6$. A possible polynomial: $\mathbf{p(x) = x^2 - x - 6}$.

Ans: A zero of a polynomial $p(x)$ is the value of $x$ at which $p(x) = 0$. Geometrically, the zeroes of $p(x)$ are the $x$-coordinates of the points where the graph of $y = p(x)$ intersects the $x$-axis. A quadratic polynomial $ax^2+bx+c$ has graph as a parabola. Since a parabola can intersect the $x$-axis at at most 2 points, a quadratic polynomial can have at most 2 zeroes.

Ans: No. A quadratic polynomial has degree 2, and by the fundamental theorem of algebra, it can have at most 2 roots. Graphically, the parabola $y = ax^2+bx+c$ can cross the $x$-axis at at most 2 points. It is impossible for a parabola to cross the $x$-axis 3 times, so a quadratic polynomial cannot have 3 zeroes.

Ans: Since the parabola opens upward, $\mathbf{a > 0}$. Since it intersects the $x$-axis at two distinct points, the discriminant $b^2 - 4ac > 0$, meaning $p(x)$ has two distinct real zeroes. The graph cuts the $x$-axis at two different points.

Ans: $x^2 - 4 = (x-2)(x+2) = 0$. Zeroes: $2$ and $-2$.

Ans: $x^2-2x-8 = (x-4)(x+2) = 0$. Zeroes: $4$ and $-2$.

Ans: $6x^2-7x-3 = (3x+1)(2x-3) = 0$. Zeroes: $\alpha = -\dfrac{1}{3}$ and $\beta = \dfrac{3}{2}$.

Ans: $4s^2-4s+1 = (2s-1)^2 = 0$. Zeroes: $\alpha = \beta = \dfrac{1}{2}$ (double zero).

Ans: $\sqrt{3}\,x^2+10x+7\sqrt{3} = (\sqrt{3}\,x+7)(x+\sqrt{3}) = 0$. Zeroes: $\alpha = -\sqrt{3}$ and $\beta = -\dfrac{7}{\sqrt{3}} = -\dfrac{7\sqrt{3}}{3}$.

Ans: $x^2-(\sqrt{3}+1)x+\sqrt{3} = (x-\sqrt{3})(x-1) = 0$. Zeroes: $\alpha = \sqrt{3}$, $\beta = 1$.

Ans: Using the quadratic formula: $x = \dfrac{-5\sqrt{2} \pm \sqrt{50+48}}{8} = \dfrac{-5\sqrt{2} \pm \sqrt{98}}{8} = \dfrac{-5\sqrt{2} \pm 7\sqrt{2}}{8}$. Zeroes: $x_1 = \dfrac{2\sqrt{2}}{8} = \mathbf{\dfrac{\sqrt{2}}{4}}$ and $x_2 = \dfrac{-12\sqrt{2}}{8} = \mathbf{-\dfrac{3\sqrt{2}}{2}}$.

Ans: $x^2+\dfrac{1}{6}x-2 = 0 \implies 6x^2+x-12 = (3x-4)(2x+3) = 0$. Zeroes: $\alpha = \dfrac{4}{3}$, $\beta = -\dfrac{3}{2}$.

Ans: $2x^2+5x-12 = (2x-3)(x+4) = 0$. Zeroes: $\alpha = \dfrac{3}{2}$, $\beta = -4$.

Ans: $abx^2+(b^2-ac)x-bc = b(ax+b)\cdot x - c(ax+b) = (ax+b)(bx-c)$. Zeroes: $x = -\dfrac{b}{a}$ and $x = \dfrac{c}{b}$. Zeroes: $-\dfrac{b}{a}$ and $\dfrac{c}{b}$.

Ans: $a=1, b=-5, c=6$.

Ans: $\alpha+\beta = -\dfrac{5}{2}$; $\alpha\beta = \dfrac{-3}{2}$. $\dfrac{1}{\alpha}+\dfrac{1}{\beta} = \dfrac{\alpha+\beta}{\alpha\beta} = \dfrac{-5/2}{-3/2} = \mathbf{\dfrac{5}{3}}$.

Ans: $\alpha+\beta = 6$; $\alpha\beta = 8$. $\alpha^2\beta+\alpha\beta^2 = \alpha\beta(\alpha+\beta) = 8\times 6 = \mathbf{48}$.

Ans: $\alpha+\beta = p$; $\alpha\beta = q$. $\alpha^2+\beta^2 = (\alpha+\beta)^2-2\alpha\beta = \mathbf{p^2-2q}$.

Ans: $\alpha+\beta=5$; $\alpha\beta=6$. $\dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha} = \dfrac{\alpha^2+\beta^2}{\alpha\beta} = \dfrac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta} = \dfrac{25-12}{6} = \mathbf{\dfrac{13}{6}}$.

Ans: $\alpha+\beta=5$; $\alpha\beta=6$. $\alpha^3+\beta^3 = (\alpha+\beta)^3-3\alpha\beta(\alpha+\beta) = 125-90 = 35$. $\dfrac{\alpha^2}{\beta}+\dfrac{\beta^2}{\alpha} = \dfrac{\alpha^3+\beta^3}{\alpha\beta} = \dfrac{35}{6} = \mathbf{\dfrac{35}{6}}$.

Ans: Zeroes: $\alpha=1, \beta=-1$. $\alpha^3+\beta^3 = 1+(-1) = \mathbf{0}$.

Ans: $\alpha+\beta = 3$; $\alpha\beta = -m(m+3)$. $\alpha^2+\beta^2 = (\alpha+\beta)^2-2\alpha\beta = 9+2m(m+3) = 9+2m^2+6m = 2m^2+6m+9$. And $(m+3)^2+m^2 = m^2+6m+9+m^2 = 2m^2+6m+9$. Hence $\alpha^2+\beta^2 = (m+3)^2+m^2$. $\blacksquare$

Ans: $\alpha+\beta=-p$; $\alpha\beta=q$. $\left(\dfrac{\alpha}{\beta}-\dfrac{\beta}{\alpha}\right)^2 = \left(\dfrac{\alpha^2-\beta^2}{\alpha\beta}\right)^2 = \dfrac{(\alpha^2-\beta^2)^2}{(\alpha\beta)^2} = \dfrac{(\alpha-\beta)^2(\alpha+\beta)^2}{q^2}$. Now $(\alpha-\beta)^2 = (\alpha+\beta)^2-4\alpha\beta = p^2-4q$. So the answer $= \dfrac{(p^2-4q)\cdot p^2}{q^2} = \mathbf{\dfrac{p^2(p^2-4q)}{q^2}}$.

Ans: Product of zeroes $= \dfrac{c}{a} = \dfrac{-6}{a} = 4 \implies a = -\dfrac{6}{4} = \mathbf{-\dfrac{3}{2}}$.

Ans: Sum of zeroes $= -\dfrac{(-k)}{3} = \dfrac{k}{3} = 3 \implies \mathbf{k = 9}$.

Ans: $\alpha+\beta=\dfrac{7}{5}$; $\alpha\beta=\dfrac{2}{5}$. $\dfrac{\alpha}{\beta^2}+\dfrac{\beta}{\alpha^2} = \dfrac{\alpha^3+\beta^3}{(\alpha\beta)^2} = \dfrac{(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)}{(\alpha\beta)^2} = \dfrac{\frac{343}{125}-3\cdot\frac{2}{5}\cdot\frac{7}{5}}{\frac{4}{25}} = \dfrac{\frac{343}{125}-\frac{42}{25}}{\frac{4}{25}} = \dfrac{\frac{343-210}{125}}{\frac{4}{25}} = \dfrac{\frac{133}{125}}{\frac{4}{25}} = \dfrac{133}{125}\times\dfrac{25}{4} = \dfrac{133}{20} = \mathbf{\dfrac{133}{20}}$.

Ans: Since $\alpha = 3$ is a zero: $9-15+k=0 \implies k=6$. Sum of zeroes $= 5$, so other zero $= 5-3 = \mathbf{2}$. $k = \mathbf{6}$.

Ans: If zeroes are $\alpha$ and $\dfrac{1}{\alpha}$, product $= \alpha\cdot\dfrac{1}{\alpha} = 1 = \dfrac{k}{5} \implies \mathbf{k = 5}$.

Ans: If zeroes are $\alpha$ and $-\alpha$, sum $= 0 = \dfrac{8k}{4} = 2k \implies \mathbf{k = 0}$.

Ans: $\alpha+\beta = 6$ and $\alpha\beta = k$. Given $3\alpha+2\beta = 20$. From $\alpha+\beta=6$: $\beta = 6-\alpha$. $3\alpha+2(6-\alpha)=20 \implies 3\alpha+12-2\alpha=20 \implies \alpha=8$, $\beta=-2$. $k = \alpha\beta = 8\times(-2) = \mathbf{-16}$.

Ans: $\alpha+\beta=-p$; $\alpha\beta=45$. $(\alpha-\beta)^2 = (\alpha+\beta)^2-4\alpha\beta = p^2-180 = 144 \implies p^2 = 324 \implies \mathbf{p = \pm 18}$.

Ans: $\alpha+\beta = -\dfrac{4}{k}$; $\alpha\beta = \dfrac{4}{k}$. $\alpha^2+\beta^2 = (\alpha+\beta)^2-2\alpha\beta = \dfrac{16}{k^2}-\dfrac{8}{k} = 24$. Let $t=\dfrac{1}{k}$: $16t^2-8t-24=0 \implies 2t^2-t-3=0 \implies (2t-3)(t+1)=0$. $t=\dfrac{3}{2}$ or $t=-1 \implies \mathbf{k=\dfrac{2}{3}}$ or $\mathbf{k=-1}$.

Ans: Let zeroes be $\alpha$ and $2\alpha$. Sum $= 3\alpha = \dfrac{k}{3} \implies \alpha = \dfrac{k}{9}$. Product $= 2\alpha^2 = \dfrac{-4}{3}$. $2\left(\dfrac{k}{9}\right)^2 = -\dfrac{4}{3} \implies \dfrac{2k^2}{81} = -\dfrac{4}{3}$. This gives $k^2<0$, which is impossible. Re-check: product $= \dfrac{c}{a} = \dfrac{-4}{3}$. $2\alpha^2 = -\dfrac{4}{3}$ gives no real $\alpha$. However, using the formula correctly: $3\alpha \cdot 2\alpha = 6\alpha^2 = -\dfrac{4}{3} \implies \alpha^2 = -\dfrac{2}{9}$ (no real solution). The problem as stated has no real solution. Note to teacher: the polynomial $3x^2-kx-4$ has product of zeroes $= \dfrac{-4}{3}$. If zeroes are $\alpha, 2\alpha$: product $= 2\alpha^2 = -\dfrac{4}{3}$ — impossible for real zeroes. A correct version: $3x^2-kx+4$ gives $2\alpha^2 = \dfrac{4}{3}$, $\alpha = \pm\sqrt{\dfrac{2}{3}}$; sum $= 3\alpha = \dfrac{k}{3}$, so $k = 9\alpha = \pm 9\sqrt{\dfrac{2}{3}} = \pm 3\sqrt{6}$.

Ans: $p(x) = x^2-3 = x^2+0\cdot x - 3$. Sum of zeroes $= -\dfrac{0}{1} = \mathbf{0}$; Product of zeroes $= \dfrac{-3}{1} = \mathbf{-3}$.

Ans: $p(x) = k[x^2-(\alpha+\beta)x+\alpha\beta] = k[x^2-5x+6]$. Taking $k=1$: $\mathbf{p(x) = x^2-5x+6}$.

Ans: $p(x) = k\!\left[x^2+\dfrac{3}{2}x-1\right]$. Taking $k=2$: $\mathbf{p(x) = 2x^2+3x-2}$.

Ans: Sum $= \sqrt{3}+(-\sqrt{3}) = 0$; Product $= \sqrt{3}\times(-\sqrt{3}) = -3$. $\mathbf{p(x) = x^2-3}$.

Ans: Sum $= 3+\sqrt{2}+3-\sqrt{2} = 6$; Product $= (3+\sqrt{2})(3-\sqrt{2}) = 9-2 = 7$. $\mathbf{p(x) = x^2-6x+7}$.

Ans: Sum $= \dfrac{1}{4}-1 = -\dfrac{3}{4}$; Product $= \dfrac{1}{4}\times(-1) = -\dfrac{1}{4}$. $p(x)=k\!\left[x^2+\dfrac{3}{4}x-\dfrac{1}{4}\right]$. Taking $k=4$: $\mathbf{p(x) = 4x^2+3x-1}$.

Ans: $\alpha+\beta=2$; $\alpha\beta=3$. New zeroes: $\alpha+2$ and $\beta+2$. New sum $= (\alpha+2)+(\beta+2) = 2+4 = 6$. New product $= (\alpha+2)(\beta+2) = \alpha\beta+2(\alpha+\beta)+4 = 3+4+4 = 11$. $\mathbf{p(x) = x^2-6x+11}$.

Ans: $\alpha+\beta=5$; $\alpha\beta=4$. New sum $= \dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha} = \dfrac{\alpha^2+\beta^2}{\alpha\beta} = \dfrac{25-8}{4} = \dfrac{17}{4}$. New product $= \dfrac{\alpha}{\beta}\cdot\dfrac{\beta}{\alpha} = 1$. $p(x)=k\!\left[x^2-\dfrac{17}{4}x+1\right]$. Taking $k=4$: $\mathbf{p(x) = 4x^2-17x+4}$.

Ans: $\alpha+\beta=-\dfrac{4}{3}$; $\alpha\beta=-\dfrac{4}{3}$. New sum $= \dfrac{1}{\alpha}+\dfrac{1}{\beta} = \dfrac{\alpha+\beta}{\alpha\beta} = 1$. New product $= \dfrac{1}{\alpha\beta} = -\dfrac{3}{4}$. $p(x)=k\!\left[x^2-x-\dfrac{3}{4}\right]$. Taking $k=4$: $\mathbf{p(x) = 4x^2-4x-3}$.

Ans: (a) – Both A and R are true and R is the correct explanation of A.

Ans: (a) – Both A and R are true and R is the correct explanation of A.

Ans: (a) – Both A and R are true and R is the correct explanation of A.

Ans: (a) – Both A and R are true and R is the correct explanation of A.

Ans: $p(x) = -x^2+5x-6 = -(x^2-5x+6) = -(x-2)(x-3)$.

Ans: $A(x) = x^2-7x+10 = (x-2)(x-5)$.

Ans: $p(x) = -2x^2+8x = -2x(x-4)$.