In the previous article, we discussed Long Answer Type Questions (Questions 1–7) of the S.N. Dey Math Exercise. In this article, we are going to solve more Long Answer Type Questions (Questions 8–15) of Quadratic Equations (Part-7) as part of our S.N. Dey Math solution series. Let's get started.
Question 8
Express the roots of the equation $q^2x^2 - (p^2 - 2q)x + 1 = 0$ in terms of those of $x^2 + px + q = 0$.
Let $\alpha, \beta$ be the roots of the equation $x^2 + px + q = 0$.
\[ \alpha + \beta = -p, \quad \alpha\beta = q \]Given equation:
\[ q^2x^2 - (p^2 - 2q)x + 1 = 0 \] \[ \implies \alpha^2\beta^2x^2 - [(\alpha + \beta)^2 - 2\alpha\beta]x + 1 = 0 \] \[ \implies \alpha^2\beta^2x^2 - (\alpha^2 + \beta^2)x + 1 = 0 \] \[ \implies (\alpha^2x - 1)(\beta^2x - 1) = 0 \] \[ \therefore x = \frac{1}{\alpha^2}, \; \frac{1}{\beta^2} \]Question 9
If the roots of the quadratics $x^2 + 2px + q = 0$ and $x^2 + 2qx + p = 0 \ (p \neq q)$ differ by a constant, show that $p + q + 1 = 0$.
Let $\alpha, \beta$ be the roots of $x^2 + 2px + q = 0$ and $\gamma, \delta$ be the roots of $x^2 + 2qx + p = 0$.
\[ \alpha + \beta = -2p, \quad \alpha\beta = q, \quad \gamma + \delta = -2q, \quad \gamma\delta = p \]According to the problem statement:
\[ |\alpha - \beta| = |\gamma - \delta| \implies (\alpha - \beta)^2 = (\gamma - \delta)^2 \] \[ \implies (\alpha + \beta)^2 - 4\alpha\beta = (\gamma + \delta)^2 - 4\gamma\delta \] \[ \implies 4p^2 - 4q = 4q^2 - 4p \] \[ \implies p^2 - q = q^2 - p \implies (p^2 - q^2) + (p - q) = 0 \] \[ \implies (p - q)(p + q) + (p - q) = 0 \implies (p - q)(p + q + 1) = 0 \]Since $p \neq q \implies p - q \neq 0$, we have:
\[ p + q + 1 = 0 \]Question 10
If one root of the equation $px^2 + qx + r = 0$ is the square of the other, prove that $q^3 + pr^2 + p^2r = 3pqr$.
Let $\alpha, \alpha^2$ be the roots of $px^2 + qx + r = 0$.
\[ \alpha + \alpha^2 = -\frac{q}{p} \quad \text{--- (1)} \] \[ \alpha \cdot \alpha^2 = \alpha^3 = \frac{r}{p} \quad \text{--- (2)} \]From (1), factoring $\alpha$:
\[ \alpha(1 + \alpha) = -\frac{q}{p} \]Cubing both sides:
\[ \alpha^3(1 + \alpha)^3 = -\frac{q^3}{p^3} \] \[ \implies \alpha^3[1 + \alpha^3 + 3\alpha(1 + \alpha)] = -\frac{q^3}{p^3} \]Substituting $\alpha^3 = \frac{r}{p}$ and $\alpha(1 + \alpha) = -\frac{q}{p}$:
\[ \frac{r}{p} \left[ 1 + \frac{r}{p} + 3\left(-\frac{q}{p}\right) \right] = -\frac{q^3}{p^3} \] \[ \implies \frac{r}{p^2} (p + r - 3q) = -\frac{q^3}{p^3} \] \[ \implies rp(p + r - 3q) = -q^3 \] \[ \implies rp^2 + r^2p - 3pqr = -q^3 \] \[ \implies q^3 + pr^2 + p^2r = 3pqr \]Question 11
If one root of the equation $ax^2 + bx + c = 0$ is the cube of the other, prove that $(b^2 - 2ca)^2 = ca(c + a)^2$.
Let $\alpha, \alpha^3$ be the roots of $ax^2 + bx + c = 0$.
\[ \alpha + \alpha^3 = -\frac{b}{a}, \quad \alpha \cdot \alpha^3 = \alpha^4 = \frac{c}{a} \]Rewriting the sum of roots:
\[ \alpha(1 + \alpha^2) = -\frac{b}{a} \]Squaring both sides:
\[ \alpha^2(1 + \alpha^2)^2 = \frac{b^2}{a^2} \] \[ \implies \alpha^2(1 + \alpha^4 + 2\alpha^2) = \frac{b^2}{a^2} \] \[ \implies \alpha^2\left(1 + \frac{c}{a}\right) + 2\alpha^4 = \frac{b^2}{a^2} \] \[ \implies \alpha^2\left(\frac{a + c}{a}\right) = \frac{b^2}{a^2} - \frac{2c}{a} = \frac{b^2 - 2ca}{a^2} \]Squaring both sides again:
\[ \alpha^4 \left(\frac{a + c}{a}\right)^2 = \left(\frac{b^2 - 2ca}{a^2}\right)^2 \]Substitute $\alpha^4 = \frac{c}{a}$:
\[ \frac{c}{a} \cdot \frac{(c + a)^2}{a^2} = \frac{(b^2 - 2ca)^2}{a^4} \] \[ \implies (b^2 - 2ca)^2 = ca(c + a)^2 \]Question 12
If the roots of the equation $ax^2 + bx + c = 0$ are two consecutive integers, prove that $b^2 - a^2 = 4ac$.
Let the consecutive integer roots be $\alpha$ and $\alpha + 1$.
\[ \alpha + (\alpha + 1) = -\frac{b}{a} \implies 2\alpha + 1 = -\frac{b}{a} \implies \alpha = -\left(\frac{a + b}{2a}\right) \quad \text{--- (1)} \] \[ \alpha(\alpha + 1) = \frac{c}{a} \quad \text{--- (2)} \]Substituting $\alpha$ from (1) into (2):
\[ -\left(\frac{a + b}{2a}\right) \left[ -\left(\frac{a + b}{2a}\right) + 1 \right] = \frac{c}{a} \] \[ \implies -\left(\frac{a + b}{2a}\right) \left(\frac{a - b}{2a}\right) = \frac{c}{a} \] \[ \implies \frac{b^2 - a^2}{4a^2} = \frac{c}{a} \] \[ \implies b^2 - a^2 = 4ac \]Question 13
If $\alpha, \beta$ are the roots of $ax^2 + 2bx + c = 0$ and $\alpha + \delta, \beta + \delta$ are those of $Ax^2 + 2Bx + C = 0$, prove that $\frac{b^2 - ac}{a^2} = \frac{B^2 - AC}{A^2}$.
For $ax^2 + 2bx + c = 0$:
\[ \alpha + \beta = -\frac{2b}{a}, \quad \alpha\beta = \frac{c}{a} \] \[ (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = \frac{4b^2}{a^2} - \frac{4c}{a} = \frac{4(b^2 - ac)}{a^2} \]For $Ax^2 + 2Bx + C = 0$:
\[ (\alpha + \delta) + (\beta + \delta) = -\frac{2B}{A}, \quad (\alpha + \delta)(\beta + \delta) = \frac{C}{A} \]Difference of roots for the second equation:
\[ [(\alpha + \delta) - (\beta + \delta)]^2 = (\alpha - \beta)^2 = \frac{4(B^2 - AC)}{A^2} \]Equating the expressions for $(\alpha - \beta)^2$:
\[ \frac{4(b^2 - ac)}{a^2} = \frac{4(B^2 - AC)}{A^2} \implies \frac{b^2 - ac}{a^2} = \frac{B^2 - AC}{A^2} \]Question 14
If the ratio of the roots of the equation $x^2 - 2px + q^2 = 0$ is equal to the ratio of the roots of $x^2 - 2rx + s^2 = 0$, prove that $p^2s^2 = q^2r^2$.
Let $\alpha, \beta$ be the roots of $x^2 - 2px + q^2 = 0 \implies \alpha + \beta = 2p, \; \alpha\beta = q^2$.
Let $\gamma, \delta$ be the roots of $x^2 - 2rx + s^2 = 0 \implies \gamma + \delta = 2r, \; \gamma\delta = s^2$.
Given $\frac{\alpha}{\beta} = \frac{\gamma}{\delta}$, applying componendo-dividendo:
\[ \frac{\alpha + \beta}{\alpha - \beta} = \frac{\gamma + \delta}{\gamma - \delta} \]Squaring both sides:
\[ \frac{(\alpha + \beta)^2}{(\alpha - \beta)^2} = \frac{(\gamma + \delta)^2}{(\gamma - \delta)^2} \] \[ \implies \frac{(\alpha + \beta)^2}{(\alpha + \beta)^2 - 4\alpha\beta} = \frac{(\gamma + \delta)^2}{(\gamma + \delta)^2 - 4\gamma\delta} \]Substituting sum and product of roots:
\[ \frac{4p^2}{4p^2 - 4q^2} = \frac{4r^2}{4r^2 - 4s^2} \implies \frac{p^2}{p^2 - q^2} = \frac{r^2}{r^2 - s^2} \] \[ \implies p^2(r^2 - s^2) = r^2(p^2 - q^2) \] \[ \implies p^2r^2 - p^2s^2 = p^2r^2 - q^2r^2 \implies p^2s^2 = q^2r^2 \]Question 15
If $\alpha$ and $\beta$ are the roots of $x^2 + px + q = 0$, show that $\frac{\alpha}{\beta}$ is a root of $qx^2 - (p^2 - 2q)x + q = 0$.
Since $\alpha, \beta$ are the roots of $x^2 + px + q = 0$, we have $\alpha + \beta = -p$ and $\alpha\beta = q$.
Consider the target equation:
\[ qx^2 - (p^2 - 2q)x + q = 0 \]Substitute $q = \alpha\beta$ and $p^2 - 2q = (-\alpha - \beta)^2 - 2\alpha\beta = \alpha^2 + \beta^2$:
\[ \alpha\beta x^2 - (\alpha^2 + \beta^2)x + \alpha\beta = 0 \] \[ \implies \alpha\beta x^2 - \alpha^2 x - \beta^2 x + \alpha\beta = 0 \] \[ \implies \alpha x(\beta x - \alpha) - \beta(\beta x - \alpha) = 0 \] \[ \implies (\beta x - \alpha)(\alpha x - \beta) = 0 \] \[ \therefore x = \frac{\alpha}{\beta}, \; \frac{\beta}{\alpha} \]Thus, $\frac{\alpha}{\beta}$ is a root of the given equation.
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