57. Quadratic Equations Solutions Part 6

1. $$D = c^2(3a^2 + b^2)^2 + 4abc^2(6a^2 + ab - 2b^2)$$

$$= c^2(9a^4 + b^4 + 6a^2b^2 + 4a^3b + 4a^2b^2 - 8ab^3)$$

$$= c^2(3a^2 - b^2 + 4ab)^2$$

which is a perfect square and hence roots are rational.

2. $$\sqrt{\frac{m}{n}} + \sqrt{\frac{n}{m}} + \frac{b}{\sqrt{ac}} = 0$$

$$L. H. S. = \sqrt{\frac{\alpha}{\beta}} + \sqrt{\frac{\beta}{\alpha}} + \frac{b}{\sqrt{ac}}$$

$$= \frac{\alpha + \beta}{\sqrt{\alpha\beta}} + \frac{b}{\sqrt{ac}}$$

$$= \frac{-\frac{b}{a}}{\sqrt{\frac{c}{a}}} + \frac{b}{\sqrt{ac}} = 0$$

3. Let $$\alpha$$ be the root, then the second root would be $$\alpha^3$$.

Product of roots $$= \alpha^4 = a \Rightarrow \alpha = a^{\frac{1}{4}}$$

Sum of roots $$= \alpha + \alpha^3 = -f(a)$$

$$\Rightarrow f(a) = -a^{\frac{1}{4}} - a^{\frac{3}{4}}$$

Therefore, the general equation in $$x$$ would be

$$f(x) = -x^{\frac{1}{4}} - x^{\frac{3}{4}}$$

4. Since $$\alpha, \beta$$ are roots of the equation $$x^2 - px + q = 0$$ therefore

$$\alpha + \beta = p$$ and $$\alpha\beta = q$$

$$(\alpha^2 - \beta^2)(\alpha^3 - \beta^3) = (\alpha - \beta)^2(\alpha + \beta)[(\alpha + \beta^2) - \alpha\beta]$$

$$=(p^2 - 4q)p(p^2 - q)$$

$$\alpha^3\beta^2 + \alpha^2\beta^3 = \alpha^2\beta^2(\alpha + \beta) = pq^2$$

Therefore, the equation would be

$$x^2 - p[(p^2 - 4q)(p^2 - q) + q^2]x + p^2q^2(p^2 - 4q)(p^2 - q) = 0$$

5. This problem has been left as an exercise.

6. Let $$\alpha, \beta$$ be the roots then $$\alpha + \beta = -\frac{b}{a}$$ and $$\alpha\beta = \frac{c}{a}$$

According to the question $$\alpha + \beta = \frac{1}{\alpha^2} + \frac{1}{\beta^2}$$

$$-\frac{b}{a} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha^2\beta^2}$$

$$-\frac{b}{a} = \frac{\frac{b^2}{a^2}}{\frac{c^2}{a^2}} - 2\frac{1}{\frac{c}{a}}$$

$$-\frac{b}{a} = \frac{b^2}{c^2} - 2\frac{a}{c}$$

$$\Rightarrow \frac{b^2}{ac} + \frac{bc}{a^2} = 2$$

7. Given, $$T = 2\pi \sqrt{\frac{h^2 + k^2}{gh}}$$

Squaring, $$h^2 + k^2 = \frac{T^2gh}{4\pi^2}$$

$$h^2 - \frac{T^2gh}{4\pi^2} + k^2 = 0$$

Clearly, $$h_1$$ and $$h_2$$ are two possible roots of above equation where

$$h_1 + h_2 = \frac{T^2g}{4\pi^2}$$ and $$h_1h_2 = k^2$$

8. Clearly, $$\alpha_1 + \alpha_2 = -p$$ and $$\alpha_1\alpha_2 = q$$

$$\beta_1 + \beta2 = -r$$ and $$\beta_1\beta_2 = s$$

Solving the two equations in $$y$$ and $$z$$ by elimination we have

$$\frac{\alpha_1}{\alpha_2} = \frac{\beta_1}{\beta_2} = k$$

$$\frac{p^2}{r^2} = \frac{(\alpha_1 + \alpha_2)^2}{(\beta_1 + \beta_2)^2}$$

$$= \frac{\alpha_1^2(1 + k^2)}{\beta_1(1 + k^2)} = \frac{\frac{\alpha_1\alpha_2}{k}}{\frac{\beta_1\beta_2}{k}} = \frac{q}{s}$$

9. $$-(1 + \alpha\beta) = -(\frac{a + c}{a})$$

1. of $$\alpha$$ and $$\beta = \frac{2\alpha\beta}{\alpha + \beta} = -\frac{2c}{b}$$

but since $$a, b, c$$ are in H. P. it becomes

$$= -\frac{2c}{\frac{2ac}{a + c}} = -(\frac{a + c}{a}) = -(1 + \alpha\beta)$$

10. Given equation is $$x + 1 = \lambda x - \lambda^2x^2$$

$$\lambda^2x^2 + (1 - \lambda)x + 1 = 0$$

$$\Rightarrow \alpha + \beta = \frac{\lambda - 1}{\lambda^2}$$ and $$\alpha\beta = \frac{1}{\lambda^2}$$

Also given that, $$\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = r - 2$$

$$\Rightarrow \alpha^2 + \beta^2 = (r - 2)\alpha\beta$$

$$(\alpha + \beta)^2 = r\alpha\beta$$

$$\frac{(\lambda - 1)^2}{\lambda^4} = \frac{r}{\lambda^2}$$

$$\Rightarrow \lambda_1 + \lambda_2 = \frac{2}{1 - r}$$ and $$\lambda_1\lambda_2 = \frac{1}{1 - r}$$

Now substituting and solving we get the desired result.

11. Let $$\alpha, \beta$$ be roots of $$ax^2 + bx + c = 0$$ then

$$\alpha + \beta = -\frac{b}{a}$$ and $$\alpha\beta = \frac{c}{a}$$

According to question, $$\frac{1}{\alpha} + \frac{1}{\beta} = -\frac{m}{l}$$ and $$\frac{1}{\alpha\beta} = \frac{n}{l}$$

From product of roots, $$\frac{c}{a} = \frac{l}{n}$$ and from sum of roots $$\frac{b}{c} = \frac{m}{l}$$

Hence the relation is established.

12. Let the roots are $$l, lm, lm^2, lm^3$$ which is an increasing G. P.

Sum of roots for first equation $$= l(1 + m) = 3$$

Sum of roots for second equation $$= lm^2(1 + m) = 12 \Rightarrow m^2 = 4 \Rightarrow m = 2$$ because G. P. is increasing.

$$\Rightarrow l = 1$$

$$A = l^2m = 2$$ and $$B = l^2m^5 = 32$$

13. For first equation, $$p + q = 2$$ and $$pq = A.$$ For second equation, $$r + s = 18$$ and $$rs = B$$

Let $$a$$ be the first term and $$d$$ be the common difference, then

$$p = a - 3d, q = a - d, r = a + d, s = a + 3d$$

Substituting in sums we have $$2a - 4d = 2$$ and $$2a + 4d = 18$$

$$\therefore a = 5$$ and $$d = 2$$

$$\therefore p = -1, q = 3, r = 7, s = 11$$

$$\therefore A = -3$$ and $$B = 77$$

14. $$\alpha + \beta = -a$$ and $$\alpha\beta = -\frac{1}{2a^2}$$

$$\alpha^4 + \beta^4 = ((\alpha + \beta)^2 - 2\alpha\beta)^2 - 2\alpha^2\beta^2$$

$$= 2 + a^4 + \frac{1}{2a^4}$$

Let $$a^4 + \frac{1}{2a^4} = y$$

$$\Rightarrow 2a^8 - 2a^4y - 1 = 0$$

Since $$a$$ is real. $$\therefore y^2 - 2 \ge 0 \Rightarrow y \ge \sqrt{2} [\because a^4 \ge 0]$$

$$\alpha^4 + \beta^4 \ge 2 + \sqrt{2}$$

15. $$\alpha + \beta = p$$ and $$\alpha\beta = q$$

$$\alpha^{\frac{1}{4}} + \beta^{\frac{1}{4}} = \sqrt[4]{\left(\alpha^{\frac{1}{4}} + \beta^{\frac{1}{4}}\right)^4}$$

$$= \sqrt[4]{\alpha + \beta + 6\sqrt{\alpha\beta} + 4\sqrt[4]{\alpha\beta(\alpha^2 + \beta^2)}}$$

$$= \sqrt[4]{p + 6\sqrt{q} + 4\sqrt[4]{q(p^2 - 2q)}}$$

16. Let $$\alpha, beta$$ be roots of first equation and $$\gamma, \delta$$ be that of second equation.

$$\alpha + \beta = \frac{b}{a}, \alpha\beta = \frac{c}{a}$$ and $$\gamma + \delta = \frac{c}{b}, \gamma\delta = \frac{a}{b}$$

According to question, $$\alpha - \beta = \gamma - \delta$$

$$(\alpha + \beta)^2 - 4\alpha\beta = (\gamma + \delta)^2 - 4\gamma\delta$$

$$\frac{b^2}{a^2} - \frac{4c}{a} = \frac{c^2}{b^2} - \frac{4a}{b}$$

$$\Rightarrow b^4 - a^2c^2 = 4ab(bc - a^2)$$

17. A cubic equation whose roots are $$\alpha, \beta, \gamma$$ is given by $$f(x) = (x - \alpha)(x - \beta)(x - \gamma)$$

$$\therefore f'(x) = (x - \alpha)(x - \beta) + (x - \beta)(x - \gamma) + (x - \alpha)(x - \gamma)$$

Now it is trivial to prove that a sign change occurs for the given limits for $$f'(x)$$ and thus a root lies in these limits.

18. Since $$\alpha, \beta, \gamma, \delta$$ are in A. P. let $$\alpha = l - 3m, \beta = l - m, \gamma = l + m, \delta = l + 3m$$ where $$l$$ is the first term and $$m$$ is the common difference of A. P.

$$\alpha + \beta = -\frac{b}{a}, \alpha\beta = \frac{c}{a}$$ and $$\gamma + \delta = -\frac{q}{p}, \gamma\delta = \frac{r}{p}$$

$$\frac{D_1}{D_2} = \frac{b^2 - 4ac}{q^2 - 4pr} = \frac{\frac{b^2}{a^2} - \frac{4c}{a}}{\frac{a^2}{p^2} - \frac{4r}{p}}\frac{a^2}{p^2}$$

$$= \frac{(\alpha - \beta)^2}{(\gamma - \delta)^2}\frac{a^2}{p^2} = \frac{4d^2}{4d^2}\frac{a^2}{p^2}$$

19. This is similar to 268 and has been left as an exercise.

20. This is a very easy problem and has been left as an exercise.

21. $$\alpha + \beta = -\frac{b}{a}, \alpha\beta = \frac{c}{a}$$ and $$\alpha^4 + \beta^4 = -\frac{m}{l}, \alpha^4\beta^4 = \frac{n}{l}$$

Discriminant of given quadratic equation, $$D = 16a^2c^2l^2 - 4a2^l(2c^2l + a^2m)$$

$$= 8a^2c^2l^2 - 4a^4lm$$

$$= 4a^4l^2\left(2\frac{c^2}{a^2} - \frac{m}{l}\right)$$

$$= 4a^4l^2(2\alpha^2\beta^2 + \alpha^4 + \beta^4) = 2a^4l^2(\alpha^2 + \beta^2)^2$$

Therefore, roots of the given equation can be computed which are found to be $$(\alpha + \beta)^2, -(\alpha + \beta)^2$$ which are equal and opposite in sign.

22. $$\alpha + \beta = -\frac{b}{a}, \alpha\beta = \frac{c}{a}$$ and $$\gamma + \delta = -\frac{m}{l}, \gamma\delta = \frac{n}{l}$$

Equation whose roots are $$\alpha\gamma + \beta\delta$$ and $$\alpha\delta + \beta\gamma$$ is

$$x^2 - (\alpha\gamma + \beta\delta + \alpha\delta + \beta\gamma)x + (\alpha\gamma + \beta\delta)(\alpha\delta + \beta\gamma) = 0$$

$$x^ - (\alpha + \beta)(\gamma + \delta)x + ((\alpha^2 + \beta^2)\gamma\delta + (\gamma^2 + \delta^2)\alpha\beta) = 0$$

$$a^2l^2x^2 - ablmx + (b^2 - 2ac)ln + (m^2 - 2ln)ac = 0$$

23. $$p$$ and $$q$$ are roots of the equation $$3x^2 - 5x - 2 = 0$$

$$\Rightarrow p + q = \frac{5}{3}$$ and $$pq = -\frac{2}{3}$$

Equation whose roots are $$3p - 2q$$ and $$3q - 2p$$ is

$$x^2 - (p + q)x - 6p^2 - 6q^2 + 13pq = 0$$

$$3x^2 - 5x - 100 = 0$$

24. Since $$p$$ and $$q$$ are roots of the equation $$x^2 + bx + c = 0$$ therefore $$p + q = -b$$ and $$pq = c$$

Equation whose roots are $$b$$ and $$c$$ is

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

$$x^2 +(p + q - pq)x - pq(p + q) = 0$$

25. Sum of roots $$= 2\alpha = -p$$ and product of roots $$= \alpha^2 - \beta = q \Rightarrow \beta = \frac{p^2 - 4q}{4}$$

Equation whose roots are $$\frac{1}{\alpha}\pm \frac{1}{\sqrt{\beta}}$$ is

$$x^2 - \frac{2}{\alpha}x + \frac{1}{\alpha^2} - \frac{1}{\beta} = 0$$

$$x^2 + \frac{2}{p}x + \frac{1}{p^2} - \frac{4}{p^2 - 4q} = 0$$

$$(p^2 - 4q)(p^2x^2 + 4px) = 16q$$

26. Proceeding like previous problem the equation can be found as $$qx^2 - p(p^2 - q)(p^2 - 4q)x - p^2q^2(p^2 - 4q) = 0$$

Rest of the problems have been left as exercises.