Differentiation (Part-38) | S N De

In the previous article, we discussed Long Answer Type Questions (Q. 8 to 15) from the S.N. Dey Mathematics Exercise. In this post, we continue with step-by-step solutions for Long Answer Type Questions (Q. 16 to 20) as part of our Quadratic Equations (Part-8) series.

Question 16

If $p^3 - q(3p - 1) + q^2 = 0$, find the relation between the roots of the equation $x^2 + px + q = 0$.

Solution

Let $\alpha, \beta$ be the roots of the quadratic equation $x^2 + px + q = 0$. Therefore:

\[ \alpha + \beta = -p \implies p = -(\alpha + \beta) \] \[ \alpha\beta = q \]

Substituting $p$ and $q$ into the given condition $p^3 - q(3p - 1) + q^2 = 0$:

\[ [-(\alpha + \beta)]^3 - \alpha\beta [-3(\alpha + \beta) - 1] + (\alpha\beta)^2 = 0 \] \[ \implies -(\alpha + \beta)^3 + 3\alpha\beta(\alpha + \beta) + \alpha\beta + \alpha^2\beta^2 = 0 \] \[ \implies -(\alpha^3 + \beta^3) + \alpha\beta + \alpha^2\beta^2 = 0 \] \[ \implies \alpha^3 + \beta^3 - \alpha\beta - \alpha^2\beta^2 = 0 \] \[ \implies \alpha(\alpha^2 - \beta) - \beta^2(\alpha^2 - \beta) = 0 \] \[ \implies (\alpha^2 - \beta)(\alpha - \beta^2) = 0 \] \[ \therefore \alpha^2 = \beta \quad \text{or} \quad \beta^2 = \alpha \]

Conclusion: One root of the equation is the square of the other root.

Question 17

Let $a, b, c$ be real numbers with $a \neq 0$ and let $\alpha, \beta$ be the roots of the equation $ax^2 + bx + c = 0$. Express the roots of $a^3x^2 + abcx + c^3 = 0$ in terms of $\alpha, \beta$.

Solution

Since $\alpha, \beta$ are the roots of $ax^2 + bx + c = 0$, we have:

\[ \alpha + \beta = -\frac{b}{a}, \quad \alpha\beta = \frac{c}{a} \]

Now consider the equation $a^3x^2 + abcx + c^3 = 0$. Divide the entire equation by $a^3$:

\[ x^2 + \left(\frac{b}{a}\right)\left(\frac{c}{a}\right)x + \left(\frac{c}{a}\right)^3 = 0 \]

Substitute $\frac{b}{a} = -(\alpha + \beta)$ and $\frac{c}{a} = \alpha\beta$:

\[ x^2 - (\alpha + \beta)(\alpha\beta)x + (\alpha\beta)^3 = 0 \] \[ \implies x^2 - (\alpha^2\beta + \alpha\beta^2)x + (\alpha^2\beta)(\alpha\beta^2) = 0 \] \[ \implies (x - \alpha^2\beta)(x - \alpha\beta^2) = 0 \]

Conclusion: The roots of $a^3x^2 + abcx + c^3 = 0$ are $\alpha^2\beta$ and $\alpha\beta^2$.

Question 18

If $\alpha$ is a root of the quadratic equation $4x^2 + 2x - 1 = 0$, prove that $4\alpha^3 - 3\alpha$ is the other root.

Solution

Let $\beta$ be the second root of $4x^2 + 2x - 1 = 0$. From the properties of roots:

\[ \alpha + \beta = -\frac{2}{4} = -\frac{1}{2} \implies \beta = -\frac{1}{2} - \alpha \quad \text{--- (1)} \]

Since $\alpha$ satisfies $4x^2 + 2x - 1 = 0$, we have:

\[ 4\alpha^2 + 2\alpha - 1 = 0 \implies 4\alpha^2 = 1 - 2\alpha \quad \text{--- (2)} \]

Now evaluate $4\alpha^3 - 3\alpha$ using equation (2):

\[ 4\alpha^3 - 3\alpha = \alpha(4\alpha^2) - 3\alpha \] \[ = \alpha(1 - 2\alpha) - 3\alpha = \alpha - 2\alpha^2 - 3\alpha = -2\alpha - 2\alpha^2 \] \[ = -2\alpha - \frac{1}{2}(4\alpha^2) = -2\alpha - \frac{1}{2}(1 - 2\alpha) \] \[ = -2\alpha - \frac{1}{2} + \alpha = -\frac{1}{2} - \alpha = \beta \quad \text{[By (1)]} \]

Conclusion: $4\alpha^3 - 3\alpha$ is indeed the other root of the equation.

Question 19

Let $\alpha, \beta$ be the roots of $x^2 - 3x + a = 0$ and $\gamma, \delta$ be the roots of $x^2 - 12x + b = 0$. If $\alpha, \beta, \gamma, \delta$ are in G.P. with a common ratio $r$, find the values of $a$ and $b$.

Solution

From the given equations:

\[ \alpha + \beta = 3, \quad \alpha\beta = a \] \[ \gamma + \delta = 12, \quad \gamma\delta = b \]

Let the roots in geometric progression be $\alpha, \beta = \alpha r, \gamma = \alpha r^2, \delta = \alpha r^3$.

Substituting into the sum of roots:

\[ \alpha + \beta = \alpha(1 + r) = 3 \quad \text{--- (1)} \] \[ \gamma + \delta = \alpha r^2(1 + r) = 12 \quad \text{--- (2)} \]

Dividing (2) by (1):

\[ \frac{\alpha r^2(1 + r)}{\alpha(1 + r)} = \frac{12}{3} \implies r^2 = 4 \implies r = \pm 2 \]

Case 1: When $r = 2$

\[ \alpha(1 + 2) = 3 \implies \alpha = 1 \] \[ \beta = 2, \quad \gamma = 4, \quad \delta = 8 \] \[ a = \alpha\beta = 1 \times 2 = 2 \] \[ b = \gamma\delta = 4 \times 8 = 32 \]

Case 2: When $r = -2$

\[ \alpha(1 - 2) = 3 \implies \alpha = -3 \] \[ \beta = 6, \quad \gamma = -12, \quad \delta = 24 \] \[ a = \alpha\beta = (-3) \times 6 = -18 \] \[ b = \gamma\delta = (-12) \times 24 = -288 \]

Question 20: Prove that the roots are real for real parameters $a, b, c$

(i) $(x - a)(x - b) = b^2$

Solution

Expanding the equation into standard quadratic form $Ax^2 + Bx + C = 0$:

\[ x^2 - (a + b)x + (ab - b^2) = 0 \]

Calculating discriminant $D = B^2 - 4AC$:

\[ D = [-(a + b)]^2 - 4(1)(ab - b^2) = a^2 + 2ab + b^2 - 4ab + 4b^2 \] \[ = a^2 - 2ab + 5b^2 = (a - b)^2 + 4b^2 \]

Since $(a - b)^2 \ge 0$ and $4b^2 \ge 0$ for all real $a, b$, $D \ge 0$.

Conclusion: The roots are always real.

(ii) $(b - c)x^2 + 2(c - a)x + (a - b) = 0$

Solution

Calculating the discriminant $D$:

\[ D = [2(c - a)]^2 - 4(b - c)(a - b) \] \[ = 4[(c - a)^2 - (ab - b^2 - ac + bc)] \] \[ = 4[c^2 - 2ca + a^2 - ab + b^2 + ac - bc] \] \[ = 2[2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca] \] \[ = 2[(a - b)^2 + (b - c)^2 + (c - a)^2] \]

Since the sum of squared terms is non-negative, $D \ge 0$.

Conclusion: The roots are always real.

(iii) $\frac{1}{x - p} + \frac{1}{x - q} = \frac{1}{a^2}$

Solution

Combining fractions and simplifying:

\[ \frac{2x - (p + q)}{(x - p)(x - q)} = \frac{1}{a^2} \implies a^2(2x - p - q) = x^2 - (p + q)x + pq \] \[ \implies x^2 - (p + q + 2a^2)x + pq + a^2(p + q) = 0 \]

Discriminant $D$:

\[ D = [-(p + q + 2a^2)]^2 - 4[pq + a^2(p + q)] \] \[ = (p + q)^2 + 4a^2(p + q) + 4a^4 - 4pq - 4a^2(p + q) \] \[ = (p + q)^2 - 4pq + 4a^4 = (p - q)^2 + (2a^2)^2 \]

Since $(p - q)^2 \ge 0$ and $(2a^2)^2 \ge 0$, $D \ge 0$.

Conclusion: The roots are always real.

(iv) $\frac{1}{x - a} + \frac{1}{x - 1} + \frac{1}{x - 2} = 0$

Solution

Multiplying through by $(x - a)(x - 1)(x - 2)$:

\[ (x - 1)(x - 2) + (x - a)(x - 2) + (x - a)(x - 1) = 0 \] \[ (x^2 - 3x + 2) + (x^2 - (a + 2)x + 2a) + (x^2 - (a + 1)x + a) = 0 \] \[ \implies 3x^2 - 2(a + 3)x + (3a + 2) = 0 \]

Discriminant $D$:

\[ D = [-2(a + 3)]^2 - 4(3)(3a + 2) = 4(a^2 + 6a + 9 - 9a - 6) \] \[ = 4(a^2 - 3a + 3) = 4\left[\left(a - \frac{3}{2}\right)^2 + \frac{3}{4}\right] \]

Since $\left(a - \frac{3}{2}\right)^2 + \frac{3}{4} > 0$ for all real $a$, $D > 0$.

Conclusion: The roots are real and distinct.