# 40. Quadratic Equations Solutions Part 1ΒΆ

For roots to be equal the discriminant has to be zero.

\(D = 4(1 + 3m)^2 - 4(1 + m)(1 + 8m) = 0\)

\(\Rightarrow 4(1 + 9m^2 + 6m - 1 - 9m -8m^2) = 0\)

\(\Rightarrow m^2 - 3m = 0 \therefore m = 0, 3\)

Discriminant of the equation is:

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

\(= 4(b^2 - 4ac)\)

Given \(a + b + c = 0 \Rightarrow b = -(a + c).\) Substituting in above equation

\(D = 4\{(a + c)^2 - 4ac\} = 4(a - c)^2 =\) a perfect square and thus roots are rational.

Discriminant of the equation is:

\(D = 4(ac + bd)^2 - 4(a^2 + b^2)(c^2 + d^2) = -4(ad - bc)^2\)

Roots are real if \(D\geq 0\) i.e. \(-4(ad - bc)^2 \geq 0 \Rightarrow (ad - bc)^2 \leq 0\)

But since \((ac - bd)^2 \nless 0 \therefore (ad - bc)^2 = 0\) i.e. \(D = 0\) (because roots are real)

Thus, if roots are real they are equal

Let \(A = a(b - c), B = b(c - a)\) and \(c = c(a - b)\)

Clearly, \(A + B + C = 0\)

Since roots are equal i.e. \(D = 0 \therefore B^2 - 4AC = 0\)

Substituting for \(B\)

\([-(A + C)^2 - 4AC] = (A - C)^2 = 0 \Rightarrow A = C \Rightarrow 2ac = ab + cb \Rightarrow b = \frac{2ac}{a + c}\)

Thus, \(a, b, c\) are in H. P.

Given equation is \((b - x)^2 - 4(a - x)(c - x) = 0\)

\(-3x^2 + 2(2a + 2c - b)x + b^2 - 4ac = 0\)

Discriminant of the above equation is:

\(D = 4(2a + 2c - b)^2 + 12(b^2 - 4ac)\)

\(= 8[(a - b)^2 + (b - c)^2 + (c - a)^2]\)

\(\because a, b, c\) are real \(\therefore D > 0\) unless \(a = b = c\)

Hence, roots are real unless \(a = b = c\)

Discriminant of the equations are \(p^2 - 4q\) and \(r^2 - 4s\)

Adding them we have \(p^2 + q^2 - 4(q + s) = p^2 + q^2 - 2pr = (p - r)^2 \geq 0\)

Thus, at least one of the discriminant is greater than zero and that equation has real roots.

Since \(x^2 - 2px + q = 0\) has equal roots \(D = 0 \Rightarrow 4p^2 - 4q = 0 \Rightarrow p^2 = q\)

Discriminant of the second equation:

\(D = 4(p + y)^2 - 4(1 + y)(q + y)\)

\(= 4[p^2 + 2y + y^2 - q -qy -y - y^2]\)

Substituting for \(q\)

\(D = -4y(p - 1)^2\)

Roots of the equation will be real and distinct only if \(D \geq 0\) but \((p - 1) \geq 0\) if \(p \neq 0\)

Thus, \(y\) has to be negative as well.

Since roots of equation \(ax^2 + 2bx + c = 0\) are equal \(\therefore 4b^2 - 4ac \geq 0\)

Discriminant of the equation \(ax^2 + 2mbx + nc = 0\) is \(4m^2b^2 - 4anc\)

Since \(m^2 > n > 0\) and \(b^ \geq ac\) \(4m^2b^2 - 4anc > 0\)

Thus, roots of the second equation are real.

Given \(ax + by = 1 \Rightarrow y = \frac{1 - ax}{b},\) substituting this in second equation

\(cx^2 + d\left(\frac{1 - ax}{b}\right)^2 = \frac{b^2cx^2 + d(1 - ax)^2}{b^2} = 1\)

\((b^2c + da^2)x^2 - 2adx + d - b^2 = 0\)

Since first two equations have one solution this equation will also have only one solution which means roots will be equal i.e. \(D = 0\)

\(\Rightarrow 4a^2d^2 - 4(b^2c + a^d)(d - b^2) = 0\)

\(b^(b^2c + a^2d - cd) = 0\)

\(\because b^ \ne 0 \therefore b^2c + a^2d - cd = 0 \Rightarrow b^2c + a^d = cd\)

Dividing both sides by \(cd\) we have

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

\(x = \frac{2ad}{2(b^2c + a^2d)} = \frac{a}{c}\)

Substituting for \(y,\) we get \(y = \frac{b}{d}\)

Let the roots of the equation be \(\alpha\) and \(r\alpha\)

Sum of roots = \(\alpha + r\alpha = -\frac{b}{a} \Rightarrow \alpha = -\frac{b}{a(r + 1)}\)

Product of roots \(= r\alpha^2 = \frac{rb^2}{a^2(1 + r)^2 = \frac{c}{a}} \Rightarrow \frac{b^2}{ac} = \frac{(r + 1)^2}{r}\)

Let the roots of the equation be \(\alpha\) and \(2\alpha.\)

Sum of roots \(= 3\alpha = -\frac{l}{l - m} \Rightarrow \alpha = -\frac{l}{l - m}\)

Product of roots \(= 2\alpha^2 = \frac{1}{l - m}\)

Substituting for \(\alpha\)

\(\frac{2l^2}{9(l - m)^2} = \frac{1}{l - m} \Rightarrow 2l^2- 9l + 9m = 0 [\because l\neq m~\text{else it would not be a quadratic equation}]\)

Since \(l\) is real, therefore discriminant of this equation would be \(\geq 0\)

\(81 - 72m \geq 0 \therefore m \leq \frac{9}{8}\)

Let the roots be \(\alpha\) and \(\alpha^n,\) then

Sum of roots \(= \alpha + \alpha^n = -\frac{b}{a}\) and product of roots \(= \alpha^{n + 1} = \frac{c}{a}\)

From products, we have \(\alpha = \left(\frac{c}{a}\right)^{\frac{1}{n + 1}}\)

From sum we have \(a\alpha^n + a\alpha + b = 0\)

Substituting value of \(\alpha\) from above

\(\Rightarrow a\left(\frac{c}{a}\right)^{\frac{n}{n + 1}} + a\left(\frac{c}{a}\right)^{\frac{1}{n + 1}} + b = 0\)

Solving this we arrive at our desired equation.

Let the roots be \(p\alpha\) and \(q\alpha.\)

Sum of roots \(= (p + q)\alpha = -\frac{b}{a}\) and product of roots \(= pq\alpha^2 = \frac{c}{a}\)

From equation for product of roots, we have \(\alpha^2 = \frac{c}{apq} \therefore \alpha = \sqrt{\frac{c}{apq}}\)

Substituting this in sum of roots and solving we arrive at desired equation.

Solutions are given below:

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

Now, \(\frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} = \frac{\alpha^3 + \beta^3}{\alpha\beta}\)

\(= \frac{(\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)}{\alpha\beta} = \frac{p(3q - p^2)}{q}\)

\((\omega\alpha + \omega^2\beta)(\omega^2\alpha + \omega\beta)\)

\(= \omega^3\alpha^2 + \omega^4\alpha\beta + \omega^2\alpha\beta + \omega^3\beta^2\)

\(= alpha^2 + \omega\alpha\beta + \omega^2\alpha\beta + \beta^2\)

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

\(= (\alpha + \beta)^2 - 3\alpha\beta = p^2 - 3q\)

Rewriting the equation we have \((A + cm^2)x^2 + Amx + Am^2 = 0\)

Sum of roots \(= \alpha + \beta = -\frac{Am}{A + cm^2}\) and product of roots \(= \alpha\beta = \frac{Am^2}{A + cm^2}\)

The expression to be evaluated is \(A(\alpha^2 + \beta^2) + A\alpha\beta + c\alpha^2\beta^2\)

\(= A[(\alpha + \beta)^2 - 2\alpha\beta] + A\alpha\beta + c(\alpha\beta)^2\)

\(= A\left[\frac{A^2m^2}{(A + cm^2)^2} - \frac{2Am^2}{A + cm^2}\right] + \frac{A^2m^2}{A + cm^2} + \frac{cA^2m^4}{(A + cm^2)^2}\)

\(= 0\)

Sum of roots \(= \alpha + \beta = -\frac{b}{a}\) and product of roots \(= \alpha\beta = \frac{c}{a}\)

Now, \(a\left(\frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha}\right) + b\left(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\right)\)

\(= \frac{a(\alpha^2 + \beta^3)}{\alpha\beta} + \frac{b(\alpha^2 + \beta^2)}{\alpha\beta}\)

\(= a\frac{[(\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)]}{\alpha\beta} + \frac{b[(\alpha + \beta)^2 - 2\alpha\beta]}{\alpha\beta}\)

Substituting for sum and product of the roots

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

Solving this we get the desired result.

Since \(a\) and \(b\) are the roots of the equation \(x^2 + px + 1 = 0\) we have \(a + b = -p\) and \(ab = 1\)

Similarly, since \(c\) and \(d\) are the roots of the equation \(x^2 + qx + 1 = 0\) we have \(c + d = -p\) and \(cd = 1\)

Now \((a - c)(b - c)(a + d)(b + d) = (ab - bc - ac + c^2)(ab + bd + ad + d^2)\)

\(= [ab - c(a + b) + c^2].[ab + d(a + b) + d^2]\)

\(= [1 + pc + c^2].[1 - pd + d^2] (\text{putting the values of } a + b~\text{and}~ab)\)

\(= 1 + cp + c^2 - pd - cdp^2 - c^2pd + d^2 + cpd^2 + c^2d^2\)

\(= 1 + (c^2 + d^2) + c^2d^2 -cdp^2 + p(c - d) + cpd(d - c)\)

\(= 1 + [(c + d)^2 - 2cd] + c^2d^2 - cdp^2 + p(c - d) + cpd(d - c)\)

Substituting for \(c + d\) and \(cd\)

\(= 1 + q^2 - 2 + 1 - p^2 + p(c - d) + p(d - c)\)

\(= q^2 - p^2\)

Let \(\alpha\) and \(\beta\) be the roots of the equation \(x^2 + px + q = 0\) then \(\alpha + \beta = -p\) and \(\alpha\beta = q.\)

Also, let \(\gamma\) and \(\delta\) be the roots of the equation \(x^2 + qx + p = 0\) then \(\gamma + \delta = -q\) and \(\gamma\delta = p\)

Now, given is that roots differ by the same quantity so we can say that

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

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

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

\(p^2 - 4q = q^2 - 4p \Rightarrow p^2 - q^2 + 4(p - q) = 0 \Rightarrow (p - q)(p + q + 4) = 0\)

Clearly, \(p \neq q\) else equations would be same \(\therefore p + q + 4 = 0\)

Since \(\alpha, \beta\) are the roots of the equation \(ax^2 + bx + c = 0\)

\(\therefore a\alpha^2 + b\alpha + c = 0\) and \(a\beta^2 + b\beta + c = 0\)

and \(\alpha + \beta = -\frac{b}{a}\) and \(\alpha\beta = \frac{c}{a}.\) Also, given \(S_n = \alpha^n + \beta^n\)

Now, \(aS_{n + 1} + bS_n + cS_{n - 1}\)

\(= a(\alpha^{n + 1} + \beta^{n + 1}) + b(\alpha^n + \beta^n) + c(\alpha^{n - 1} + \beta^{n - 1})\)

\(= \alpha^{n - 1}(a\alpha^2 + b\alpha + c) + \beta^{n - 1}(a\beta^2 + b\beta + c)\)

\(= \alpha^{n - 1}.0 + \beta^{n - 1}.0\)

\(\therefore S_{n + 1} = -\frac{b}{a}S_n -\frac{c}{a}S_{n - 1}\)

Substituting \(n = 4\) we have

\(S_5 = -\frac{b}{a}S_4 - \frac{c}{a}S_3\)

\(= -\frac{b}{a}(-\frac{b}{a}S_3 - \frac{c}{a}S-2) - \frac{c}{a}S_3\)

\(= \left(\frac{b^2}{a^2} - \frac{c}{a}\right)S_3 + \frac{bc}{a^2}S_2\)

Proceeding similarly we have the solution as

\(= -\frac{b}{a^6}(b^2 - 2ac)^2 + \frac{(b^2 - ac)bc}{a^4}\)

Let \(\alpha\) and \(\beta\) be the roots of the equation \(ax^2 + bx + c = 0\)

Given, \(\alpha + \beta = \frac{1}{\alpha^2} + \frac{1}{\beta^2}\)

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

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

\(-bc^2 = ab^2 - 2a^2c \Rightarrow ca^2 = \frac{ab^2 + bc^2}{2}\)

Thus, \(bc^2, ca^2, ab^2\) are in A. P.

Rewriting the equation \(m^2x^2 + (2m - m^2)x + 3 = 0\)

Since \(\alpha\) and \(\beta\) are the roots of the equation \(\alpha + \beta = -\frac{2m - m^2}{m^2} = \frac{m - 2}{m}\) and \(\alpha\beta = \frac{3}{m^2}\)

Given, \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{4}{3} \Rightarrow \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{4}{3}\)

\(3(\alpha^2 + \beta^2) = 4\alpha\beta \Rightarrow 3[(\alpha + \beta)^2 - 2\alpha\beta] = 4\alpha\beta\)

\(3(\alpha + \beta)^2 - 10\alpha\beta = 0 \Rightarrow 3\left[\left(\frac{m - 2}{m}\right)^2 - \frac{10}{m^2}\right] = 0\)

\(m^2 - 4m - 6 = 0\)

Since \(m_1, m_2\) are two values of \(m\) we have \(m_1 + m_2 = 4\) and \(m_1m_2 = -6\)

Now, \(\frac{m_1^2}{m_2} + \frac{m_2^2}{m_1} = \frac{m_1^3 + m_2^3}{m_1m_2}\)

\(= \frac{(m_1 + m_2)^3 - 3m_1m_2(m_1 + m_2)}{3m_1m_2} = -\frac{68}{3}\)

Let \(\alpha\) and \(\beta\) be the roots of the equation \(ax^2 + bx + c = 0;\) \(\gamma\) and \(\delta\) are the roots of the equation \(a_1x^2 + b_1x + c_1 = 0,\) then

\(\alpha + \beta = -\frac{b}{a}, \alpha\beta = \frac{c}{a}\) and \(\gamma + \delta = -\frac{b_1}{a_1}, \gamma\delta = \frac{c_1}{a_1}\)

According to question, \(\frac{\alpha}{\beta} = \frac{\gamma}{\delta}\)

By componendo and dividendo,

\(\frac{\alpha - \beta}{\alpha + \beta} = \frac{\gamma - \delta}{\gamma + \delta}\)

Squaring both sides

\(\left(\frac{\alpha - \beta}{\alpha + \beta}\right)^2 = \left(\frac{\gamma - \delta}{\gamma + \delta}\right)^2\)

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

\(\frac{b^2 - 4ac}{b^2} = \frac{b_1^2 - 4a_1c_1}{b_1^2} \Rightarrow -4acb_1^2 = -4a_1c_1b^2 \Rightarrow \left(\frac{b}{b_1}\right)^2 = \frac{ac}{a_1c_1}\)

Since irrational roots appear in pairs and are conjugate. Thus, if first root is \(\alpha = \frac{1}{2 + \sqrt{5}}\)

\(\alpha = \frac{1}{2 + \sqrt{5}}\frac{2 - \sqrt{5}}{2 - \sqrt{5}} = \frac{2 - \sqrt{5}}{4 - 5} = -2 + \sqrt{5}\)

Then second root would be \(\beta = -2 + \sqrt{5}\)

\(\alpha + \beta = -4\) and \(\alpha\beta = -1\)

Therefore, the equation is \(x^2 - (\alpha + \beta)x + \alpha\beta = 0 \Rightarrow x^2 + 4x -1 = 0\)

Since \(\alpha\) and \(\beta\) are the roots of the equation \(\therefore \alpha + \beta = -\frac{b}{a}\) and \(\alpha\beta = \frac{c}{a}\)

Sum of the roots for which quadratic equation is to be found \(= \frac{1}{a\alpha + b} + \frac{1}{a\beta + b}\)

\(= \frac{a(\alpha + \beta) + 2b}{a^2\alpha\beta + ab(\alpha + \beta) + b^2} = \frac{a\left(-\frac{b}{a}\right) + 2b}{a^2.\frac{c}{a} + av\left(-\frac{b}{a}\right)} + b^2\)

\(= \frac{b}{ac}\)

Product of the roots \(= \left(\frac{1}{a\alpha + b}\right)\left(\frac{1}{a\beta + b}\right)\)

\(= \frac{1}{a^2\alpha\beta + ab(\alpha + \beta) + b^2} = \frac{1}{a^2.\frac{c}{a} + ab\left(-\frac{c}{a}\right) + b^2} = \frac{1}{ac}\)

Therefore, the equation is \(x^2 - \frac{b}{ac}x + \frac{1}{ac} = 0 \Rightarrow acx^2 - bx + 1 = 0\)

Given equation is \((x - a)(x - b) - k = 0 \Rightarrow x^2 - (a + b)x + ab - k = 0\)

Since \(c, d\) are roots of this equation \(\Rightarrow c + d = a + b\) and \(cd = ab - k\)

The equation where roots are \(a, b\) is \(x^2 - (a + b)x + ab = 0 \Rightarrow x^2 - (c + d)x + cd + k = 0\)

Correct equation is \(x^2 + 13x + q = 0\) and incorrect equation is \(x^ + 17x + q = 0\)

Roots of correct incorrect equation are \(-2\) and \(-15.\) Thus \(q = 30\)

Therefore, correct equation is \(x^2 + 13x + 30 = 0\) and thus roots are \(-3, -10.\)

Clearly, \(\alpha + \beta = -p\) and \(\alpha\beta = q\)

Substituting \(x = \frac{\alpha}{\beta}\) in the given equation we have

\(q\frac{\alpha^2}{\beta^2} - (p^2 - 2q)\frac{\alpha}{\beta} + q = 0\)

\(\Rightarrow q\alpha^2 - (p^2 - 2q)\alpha\beta + q\beta^2 = 0\)

\(q(\alpha^2 + \beta^2) - (p^2 - 2q)q = 0\)

\(q[(\alpha + \beta)^2 - 2\alpha\beta] - (p^2 - 2q)q = 0\)

\(q(p^2 - 2q) - (p^2 - 2q)q = 0 \Rightarrow 0 = 0\)

Thus, \(\frac{\alpha}{\beta}\) is a root of the given equation.

Let \(\alpha\) and \(\beta\) be the roots of \(x^2 - ax + b = 0\) and \(\alpha\) be the common and equal root from the second equation \(x^2 - px + q = 0\)

Thus, \(\alpha + \beta = a, \alpha\beta = b\) and \(2\alpha = p, \alpha^2 = q\)

\(b + q = \alpha\beta + \alpha^2 = \alpha(\beta + \alpha) = \frac{p}{2}a = \frac{ap}{2}\)

Let \(\alpha\) be the common root. Then, we have

\(a\alpha^2 + 2b\alpha + c = 0\) and \(a_1\alpha^2 + 2b_1\alpha + c_1 = 0\)

Solving equations by cross-multiplication we have

\(\frac{\alpha^2}{2(bc_1 - b_1c)} = \frac{\alpha}{(ca_1 - a_1c)} = \frac{1}{2(ab_1 - a_1b)}\)

From first two we have \(\alpha\) as

\(\alpha = \frac{2(bc_1 - b_1c)}{ca_1 - a_1c}\)

and from last two we have \(\alpha\) as

\(\alpha = \frac{ca_1 - ac_1}{2(ab_1 - a_1b)}\)

Equating we get

\(\frac{2(bc_1 - b_1c)}{ca_1 - a_1c} = \frac{ca_1 - ac_1}{2(ab_1 - a_1b)}\)

\((ca_1 - ac_1)^2 = 4(ab_1 - a_1b)(bc_1 - b_1c)\)

Given, \(\frac{a}{a_1}, \frac{b}{b_1}, \frac{c}{c_1}\) are in A. P., let \(d\) be the common difference.

\(\left(\frac{c}{c_1} - \frac{a}{a_1}\right)^2c_1^2a_1^2 = 4\left(\frac{a}{a_1} - \frac{b}{b_1}\right)a_1b_2\left(\frac{b}{b_1} - \frac{c}{c_1}\right)b_1c_1\)

\((2d)^2c_1^2a_2^2 = 4(-d)a_1b_1(-d)b_1c_1\)

\(4d^2c_1^2a_1^2 = 4d^2a_1c_1b_1^2 \Rightarrow c_1a_1 = b_1^2\)

Thus, \(a_1, b_1, c_1\) are in G. P.

Let \(\alpha\) be the common root between first two, \(\beta\) be the common root between last two and \(\gamma\) be the common root between first and last equations.

Thus, \(\alpha\) and \(\beta\) are the roots of the first equation.

\(\Rightarrow \alpha + \gamma = -p_1, \alpha\gamma = q_1\)

Similarly,

\(\alpha + \beta = -p_2, \alpha\beta = q_2\)

\(\beta + \gamma = -p_3, \beta\gamma = q_3\)

\(L. H. S. = (p_1 + p_2 + p_3)^2 = 4(\alpha + \beta + \gamma)^2\)

\(R. H. S. = 4(p_1p_2 + p_2p_3 + p_1p_3 - q_1 - q_2 - q_3)\)

\(= 4[(\alpha + \gamma)(\alpha + \beta) + (\alpha + \beta)(\beta + \gamma) + (\alpha + \gamma)(\beta + \gamma) - \alpha\gamma - \alpha\beta - \beta\gamma]\)

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

Hence, proven that \(L. H. S. = R. H. S.\)

Let \(\alpha\) be the common root then we have

\(\alpha^2 + c\alpha + ab = 0\) and \(\alpha^2 + b\alpha + ca = 0\)

By cross-multiplication, we get the solution as

\(\frac{\alpha^2}{ac^2 - ab^2} = \frac{\alpha}{ab - ac} = \frac{1}{b - c}\)

From first two we have \(\alpha = \frac{ac^2 - ab^2}{ab - ac} = -(b + c)\)

From last two we have \(\alpha = a\)

Equating these two we get \(a = -(b + c) \Rightarrow a + b + c = 0\)

Let the other root of the equations be \(\beta\) and \(\beta1\) then we have

\(\alpha\beta = ab\) and \(\alpha\beta1 = ca\)

\(\therefore \beta = b\) and \(\beta1 = c\)

Equation whose roots are \(\beta\) and \(\beta1\) is

\(x^2 - (\beta + \beta1)x + \beta\beta1 = 0 \Rightarrow x^2 -(b + c) + bc = 0 \Rightarrow x^2 + ax + c = 0\)

Clearly, root of the equation \(x^2 + 2x + 9 = 0\) are imaginary and since they appear in pairs both the roots will be common and thus the ratio of the coefficients of the terms will be equal.

\(a : b: c = 1 : 2 : 9\)

Since both the equations have only one common root so the roots must be rational as irrational and complex roots appear in pairs. Thus, the roots of these two equations must be rational and therefore the discriminants must be perfect squares.

Therefore, \(b^2 - ac\) and \(b_1^2 - a_1c_2\) must be perfect squares.

Let \(\alpha\) be a common root. Then, we have

\(3\alpha^2 -2\alpha + p = 0\) and \(6\alpha^2 - 17\alpha + 12 = 0\)

Solving by cross-multiplication

\(\frac{\alpha^2}{-24 + 17p} = \frac{\alpha}{6p - 36} = \frac{1}{-39}\)

From first two we have \(\alpha = \frac{17p - 24}{6p - 36}\)

and from last two we have \(\alpha = \frac{6p - 36}{-39} = -\frac{2p - 12}{13}\)

Equating these two and solving for \(p\) we get \(p = -\frac{15}{4}, -\frac{8}{3}.\)

When \(x = 0, |x|^2 - |x| - 2 = |0|^2 - |0| - 2 = -2 \ne 0\)

Since it is not satisfied by \(x = 0\) it is an equation.

When \(x = -a\) the equation is satisfied. Similarly, it is satisfied by values of \(x\) being \(-b\) and \(-c\). The highest power of \(x\) occurring is \(2\) and is true for three distinct values of \(x\) therefore it cannot be equation but an identity.

Equating the coefficients for similar powers of \(x\) we get

Coefficient of \(x^2:\) \(a^2 - 1 = 0 \Rightarrow a = \pm1\)

Coefficient of \(x:\) \(a - 1 = 0 \Rightarrow a = 1\)

Constant term: \(a^2 - 4a + 3 = 0 \Rightarrow a = 1, 3\)

The common value of \(a\) is 1 which will make this an identity.

Given, \(\left(x + \frac{1}{c}\right)^2 = 4 + \frac{3}{2}\left(x - \frac{1}{x}\right)\)

\(\left(x + \frac{1}{x}\right)^2 - 4 - \frac{3}{2}\left(x - \frac{1}{x}\right) = 0\)

\(\left\{\left(x - \frac{1}{x}\right)^2 + 4x\frac{1}{x}\right\} - \frac{3}{2}\left(x - \frac{1}{x}\right) - 4 = 0\)

Substituting \(a = x - \frac{1}{x}\)

\(a^2 - \frac{3}{2}a = 0 \Rightarrow 2a^2 - 3a = 0 \therefore a = 0, \frac{3}{2}\)

\(x - \frac{1}{x} = 0 \Rightarrow x = \pm1\)

\(x - \frac{1}{x} - \frac{3}{2} \Rightarrow x = 2, -\frac{1}{2}\)

Given equation is \((x + 4)(x + 7)(x + 8)(x + 11) + 20 = 0\)

Rewriting the equation,

\([(x + 4)(x + 11)][(x + 7)(x + 8)] + 20 = 0\)

\((x^2 + 15x + 44)(x^2 + 15x + 56) + 20 = 0\)

Substituting \(a = x^2 + 15x,\) we get

\((a + 44)(a + 56) + 20 = 0\)

\(\Rightarrow a = -46, -54\)

If \(a = -46 \Rightarrow x^2 + 15x + 46 = 0 \Rightarrow x = \frac{-15 \pm \sqrt{41}}{2}\)

If \(a = -54 \Rightarrow x^2 + 15x + 54 = 0 \Rightarrow x = - 6, -9\)

Given equation is \(3^{2x + 1} + 3^2 = 3^{x + 3} + 3^x\)

Let \(3^x = a,\) then we have

\(3a^2 + 9 = 28a \Rightarrow 3a^2 - 28a + 9 = 0\)

\(a = \frac{1}{3}, 9\)

If \(a = \frac{1}{3} \Rightarrow x = -1\)

If \(a = 9 \Rightarrow x = 2\)

Clearly, \((5 + 2\sqrt{6})^{x^2 - 3}(5 - 2\sqrt{6})^{x^2 - 3} = 1\)

Let \((5 + 2\sqrt{6})^{x^2 - 3} = 1\) then \((5 - 2\sqrt{6})^{x^2 - 3} = \frac{1}{y}\)

The given equation becomes \(y + \frac{1}{y} = 10\) where \(y = (5 + 2\sqrt{6})^{x^2 - 3}\)

\(\Rightarrow y^2 -10y + 1 = 0\)

Solving the equation we have roots as \(y = 5 \pm 2\sqrt{6}\)

\(\therefore x^2 - 3 = \pm 1\)

\(x = \pm2, \pm\sqrt{2}\)

Let the speed of the bus \(= x\) km/hour \(\therefore\) the speed of car \(= x + 25\) km/hour.

Time taken by bus \(= \frac{500}{x}\) hours and by car \(= \frac{500}{x + 25}\) hours

Given, \(\frac{500}{x} = \frac{500}{x + 25} + 10\)

\(\Rightarrow x^2 - 25x + 1250 = 0\)

\(x = -50, 25\) but \(x\) cannot be negative as it is a scalar quantity. Thus, speed of car = \(50\) km/hour.

Given equation is \((a + b)^2x^2 - 2(a^2 - b^2)x + (a - b)^2 = 0\)

Discriminant \(= 4(a^2 - b^2)^2 - 4(a + b)^2(a - b)^2 = 0\)

Since discriminant is zero, roots are equal.

Given equation is \(3x^2 + 7x + 8 = 0\)

Discriminant \(D = 49 - 96 < 0\)

Since it is negative roots will be complex and conjugate pair.

Given equation is \(3x^2 + (7 + a) + 8 - a = 0\)

Discriminant \(D = (7 + a)^2 + 12a\)

For roots to be equal it has to be zero.

\(\Rightarrow a^2 + 26a + 49 = 0\)

\(\Rightarrow a = 13 \pm 6\sqrt{6}\)

It is given that roots are equal i.e. discriminant is zero.

\(\Rightarrow 4(ac + bd)^2 - 4(a^2 + b^2)(c^2 + d^2) = 0\)

\(a^2c^2 + b^2d^2 - 2abcd - a^2c^2 - a^2d^2 - b^2c^2 - b^2d^2 = 0\)

\((ad - bc)^2 = 0\)

\(ad = bc \Rightarrow \frac{a}{b} = \frac{c}{d}\)

Discriminant is \(4(c - a)^2 - 4(b - c)(a - b)\)

\(= c^2 + a^2 -2ac - ab + b^2 + ac - bc\)

\(= a^2 + b^2 + c^2 - ab - bc - ac\)

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

Clearly the above expression is either greater than zero or equal to zero. Hence, roots are real.

Given equation is

\(x^2 - x + x^2 - (a + 1)x + a + x^2 - ax = 0\)

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

Discriminant \(D = 4(a + 1)^2 - 12a\)

\(= a^2 + 2a + 1 - 3a = a^2 - a + 1 = (a - 1)^2 + a\)

which is greater than zero for all \(a\) and hence roots are real.

Discriminant of the equation \(D = b^2 - 4ac\)

Given, \(a + b + c = 0 \Rightarrow b = -(a + c)\)

Substituting value of \(b\)

\(D = (a + c)^2 - 4ac = (a - c)^2\)

which is either zero or positive. Hence, roots are rational.

This has been left as an exercise to the reader.