# 50. Quadratic Equations Solutions Part 6ΒΆ

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

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

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

which is a perfect square and hence roots are rational.

\(\sqrt{\frac{m}{n}} + \sqrt{\frac{n}{m}} + \frac{b}{\sqrt{ac}} = 0\)

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

\(= \frac{\alpha + \beta}{\sqrt{\alpha\beta}} + \frac{b}{\sqrt{ac}}\)

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

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

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

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

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

This problem has been left as an exercise.

Let \(\alpha, \beta\) be the roots then \(\alpha + \beta = -\frac{b}{a}\) and \(\alpha\beta = \frac{c}{a}\)

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

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

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

\(-\frac{b}{a} = \frac{b^2}{c^2} - 2\frac{a}{c}\)

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

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 \(h_1h_2 = k^2\)

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

\(\frac{\alpha_1}{\alpha_2} = \frac{\beta_1}{\beta_2} = k\)

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

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

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

- of \(\alpha\) and \(\beta = \frac{2\alpha\beta}{\alpha + \beta} = -\frac{2c}{b}\)

but since \(a, b, c\) are in H. P. it becomes

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

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.

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.

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 \(B = l^2m^5 = 32\)

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 \(B = 77\)

\(\alpha + \beta = -a\) 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. \(\therefore y^2 - 2 \ge 0 \Rightarrow y \ge \sqrt{2} [\because a^4 \ge 0]\)

\(\alpha^4 + \beta^4 \ge 2 + \sqrt{2}\)

\(\alpha + \beta = p\) and \(\alpha\beta = q\)

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

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

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

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, \(\alpha - \beta = \gamma - \delta\)

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

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

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

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.

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 \(\gamma + \delta = -\frac{q}{p}, \gamma\delta = \frac{r}{p}\)

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

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

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

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

\(\alpha + \beta = -\frac{b}{a}, \alpha\beta = \frac{c}{a}\) 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.

\(\alpha + \beta = -\frac{b}{a}, \alpha\beta = \frac{c}{a}\) 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

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

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

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

\(p\) 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

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

\(3x^2 - 5x - 100 = 0\)

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

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

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

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

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

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

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

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.