Quadratic Equation MCQ Questions & Answers in Algebra | Maths

Let $$\alpha $$ be the common root to the equations :
$$\eqalign{ & {x^2} - kx - 21 = 0\,\,\,{\text{and }}{x^2} - 3kx + 35 = 0 \cr & \therefore '\alpha '{\text{ satisfies both the equations}} \cr & \therefore {\alpha ^2} - k\alpha - 21 = 0\,\,\,\,\,\,.....\left( {\text{i}} \right) \cr & {\text{and }}{\alpha ^2} - 3k\alpha + 35 = 0\,\,\,\,.....\left( {{\text{ii}}} \right) \cr & {\text{From }}\left( {\text{i}} \right)\,\,{\text{and }}\left( {{\text{ii}}} \right), \cr & {\alpha ^2} - 21 = \frac{{{\alpha ^2} + 35}}{3} \cr & \Rightarrow 3{\alpha ^2} - 63 = {\alpha ^2} + 35 \cr & \Rightarrow {\alpha ^2} = 49 \cr & \Rightarrow \alpha = \pm 7 \cr} $$
Now, again by eliminating $${\alpha ^2}$$ from (i) and (ii), we get
$$\eqalign{ & k\alpha + 21 = 3k\alpha - 35 \cr & \Rightarrow 2k\alpha = 56 \cr & \Rightarrow k = \frac{{56}}{{2\alpha }} \cr & {\text{When, }}\alpha = 7{\text{ then }}k = 4 \cr & {\text{When, }}\alpha = - 7{\text{ then }}k = - 4 \cr & {\text{Hence, }}k = \pm 4 \cr} $$

$$\eqalign{ & \alpha > 0,\beta > 0,{D_1} \geqslant 0 \cr & \Rightarrow \,\,\alpha + \beta > 0,\alpha \beta > 0,{D_1} \geqslant 0. \cr & \gamma < 0,\delta < 0,{D_2} \geqslant 0 \cr & \Rightarrow \,\,\gamma + \delta < 0,\gamma \delta > 0,{D_2} \geqslant 0. \cr & \therefore \,\,a - 1 > 0,3 > 0,{\left( {a - 1} \right)^2} - 12 \geqslant 0. \cr & {\text{and }} - 3 < 0,6 - a > 0,9 - 4\left( {6 - a} \right) \geqslant 0. \cr} $$
Solve the inequations together.

Given quadratic equation is $$ax^2 + bx + c = 0$$    whose one root is $$\frac{1}{{2 - \sqrt { - 2} }}$$
Consider $$\frac{1}{{2 - \sqrt { - 2} }} = \frac{1}{{2 - \sqrt 2 i}} \times \frac{{2 + \sqrt 2 i}}{{2 + \sqrt 2 i}}$$
$$ = \frac{{2 + \sqrt 2 i}}{{4 + 2}} = \frac{{2 + \sqrt 2 i}}{6}$$
∴ Another root will be $$\frac{{2 - \sqrt 2 i}}{6}$$
( ∵ complex roots always occurs in pairs )
Thus, sum of roots $$ = \frac{{2 + \sqrt 2 i}}{6} + \frac{{2 - 2\sqrt 2 i}}{6} = \frac{4}{6}$$
and product of roots $$ = \left( {\frac{{2 + \sqrt 2 i}}{6}} \right)\left( {\frac{{2 - \sqrt 2 i}}{6}} \right) = \frac{{4 + 2}}{{36}} = \frac{1}{6}$$
∴ Required equation is
$$\eqalign{ & {x^2} - \left( {{\text{sum of roots}}} \right)x + \left( {{\text{product of roots}}} \right) = 0 \cr & {x^2} - \frac{4}{6}x + \frac{1}{6} = 0 \cr & \Rightarrow 6{x^2} - 4x + 1 = 0 \cr} $$
Thus, the values of $$a, b, c$$  are $$6, - 4, 1$$  respectively.

$$\eqalign{ & \alpha ,\beta \,\,{\text{are roots of }}{x^2} - x + 1 = 0 \cr & \therefore \,\,\alpha = - \omega \,\,{\text{and }}\beta = - {\omega ^2} \cr} $$
where $$\omega $$ is cube root of unity
$$\eqalign{ & \therefore \,\,{\alpha ^{101}} + {\beta ^{107}} = {\left( { - \omega } \right)^{101}} + {\left( { - \omega } \right)^{107}} \cr & = - \left[ {{\omega ^2} + \omega } \right] = - \left[ { - 1} \right] = 1 \cr} $$

$$\eqalign{ & {x^2} - 2{p_1}x + 1 = 0\,\,{\text{has roots }}\alpha ,\beta \cr & {x^2} - 4{p_2}x + 2 = 0\,\,{\text{has roots }}\alpha ,\gamma \cr & {x^2} - 6{p_3}x + 3 = 0\,\,{\text{has roots }}\beta ,\gamma \cr & \Rightarrow \,\,\alpha + \beta = 2{p_1},\alpha + \gamma = 4{p_2},\beta + \gamma = 6{p_3},\alpha \beta = 1,\alpha \gamma = 2,\beta \gamma = 3. \cr & \therefore \,\,\gamma - \beta = 4{p_2} - 2{p_1}\,\,{\text{and }}\alpha \left( {\gamma - \beta } \right) = 1 \cr & \Rightarrow \,\,\alpha = \frac{1}{{4{p_2} - 2{p_1}}}. \cr & {\text{Also }}\frac{\beta }{\gamma } = \frac{1}{2} \cr & \therefore \,\,2\beta = \gamma \,\,\,\therefore \,\,\beta \cdot 2\beta = 3\,\,\,\,\therefore \,\,\beta = \pm \sqrt {\frac{3}{2}} \cr & \therefore \,\,\frac{1}{{4{p_2} - 2{p_1}}}\left( { \pm \sqrt {\frac{3}{2}} } \right) = 1\,\,\,\,\,\,.....\left( 1 \right) \cr & {\text{Again }}\frac{1}{\alpha } + \frac{1}{\beta } = 2{p_1},\frac{1}{\alpha } + \frac{1}{\gamma } = 2{p_2},\frac{1}{\beta } + \frac{1}{\gamma } = 2{p_3} \cr & \therefore \,\,\frac{1}{\alpha } + \frac{1}{\beta } + \frac{1}{\gamma } = {p_1} + {p_2} + {p_3} \cr & \therefore \,\,\frac{1}{\alpha } = {p_1} + {p_2} - {p_3},\frac{1}{\beta } = {p_1} - {p_2} + {p_3},\frac{1}{\gamma } = - {p_1} + {p_2} + {p_3}. \cr & {\text{So, }}\frac{\gamma }{\beta } = 2 = \frac{{{p_1} - {p_2} + {p_3}}}{{ - {p_1} + {p_2} + {p_3}}}\,{\text{and }}\frac{\gamma }{\alpha } = 3 = \frac{{ - {p_1} + {p_2} + {p_3}}}{{{p_1} + {p_2} - {p_3}}}. \cr} $$
Thus, we get $$\frac{{{p_1}}}{7} = \frac{{{p_2}}}{4} = \frac{{{p_3}}}{9} = t.$$     Using this in (1), we get two values of $$t.$$

Let $$p, q, r$$  are in A.P.
$$ \Rightarrow \,\,2q = p + r\,\,\,\,\,\,\,\,\,......\left( {\text{i}} \right)$$
$$\eqalign{ & {\text{Given }}\frac{1}{\alpha } + \frac{1}{\beta } = 4 \cr & \Rightarrow \,\,\frac{{\alpha + \beta }}{{\alpha \beta }} = 4 \cr & {\text{We have }}\alpha + \beta = - \frac{q}{p}\,\,{\text{and }}\alpha \beta = \frac{r}{p} \cr & \Rightarrow \,\,\frac{{ - \frac{q}{p}}}{{\frac{r}{p}}} = 4 \cr & \Rightarrow \,\,q = - 4r\,\,\,\,......\left( {{\text{ii}}} \right) \cr & {\text{From }}\left( {\text{i}} \right),{\text{ we have}} \cr & {\text{2}}\left( { - 4r} \right) = p + r \cr & \Rightarrow \,\,p = - 9r \cr & q = - 4r \cr & {\text{Now }}\left| {\alpha - \beta } \right| = \sqrt {{{\left( {\alpha + \beta } \right)}^2} - 4\alpha \beta } \cr & = \sqrt {{{\left( {\frac{{ - q}}{p}} \right)}^2} - \frac{{4r}}{p}} = \frac{{\sqrt {{q^2} - 4pr} }}{{\left| p \right|}} \cr & = \frac{{\sqrt {16{r^2} + 36{r^2}} }}{{\left| { - 9r} \right|}} = \frac{{2\sqrt {13} }}{9} \cr} $$

$$\eqalign{ & {b^2} - 4ac < 0\,\,{\text{and }}a + c < b. \cr & {b^2} - 4ac < 0 \cr} $$
⇒ $$a, c$$  must be of the same sign.
If $$a, c$$  are both positive, $${\left( {a + c} \right)^2} < {b^2}.$$
$$\eqalign{ & \therefore \,\,{b^2} - 4ac < 0 \cr & \Rightarrow \,\,{\left( {a + c} \right)^2} - 4ac < 0 \cr & \Rightarrow \,\,{\left( {a - c} \right)^2} < 0\,\,{\text{which is absurd}}{\text{.}} \cr} $$
So $$a, c$$  will both be negative. For example, $$ - {x^2} + 4x - 5 = 0.$$
It satisfies, $$D < 0$$  and $$a + c < b.$$  Also $$4a + c = - 4 - 5 = - 9 < 8 = 2b.$$
Clearly, when $$a, c$$  are negative, $$b$$ must be positive. Otherwise, the equation becomes one where co-efficients are all positive.

$$\eqalign{ & D = {b^2} - 4ac,D' = {d^2} + 4ac \cr & \Rightarrow \,\,D + D' = {b^2} + {d^2} > 0 \cr} $$
∴ at least one of $$D, D'$$  is positive.

KEY CONCEPT:
If $$f\left( \alpha \right)$$ and $$f\left( \beta \right)$$ are of opposite signs then there must lie a value $$\gamma $$ between $$\alpha $$ and $$\beta $$ such that $$f\left( \gamma \right) = 0.$$
$$a,b,c$$   are real numbers and $$a \ne 0.$$
$$\eqalign{ & {\text{As }}\,\alpha \,\,{\text{is a root of }}\,{a^2}{x^2} + bx + c = 0 \cr & \therefore \,\,{a^2}{\alpha ^2} + b\alpha + c = 0\,\,\,\,\,\,\,\,\,.....\left( 1 \right) \cr & {\text{Also }}\beta \,{\text{ is root of }}\,{a^2}{x^2} - bx - c = 0 \cr & \therefore \,\,{a^2}{\beta ^2} - b\beta - c = 0\,\,\,\,\,\,\,\,.....\left( 2 \right) \cr & {\text{Now, let}}\,\,f\left( x \right) = {a^2}{x^2} + 2bx + 2c \cr & {\text{Then}}\,\,f\left( \alpha \right) = {a^2}{\alpha ^2} + 2b\alpha + 2c \cr & = {a^2}{\alpha ^2} + 2\left( {b\alpha + c} \right) \cr & = {a^2}{\alpha ^2} + 2\left( { - {a^2}{\alpha ^2}} \right)\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{Using eq}}{\text{.}}\left( 1 \right)} \right] \cr & = - {a^2}{\alpha ^2}. \cr & {\text{and }}\,f\left( \beta \right) = {a^2}{\beta ^2} + 2b\beta + 2c \cr & = {a^2}{\beta ^2} + 2\left( {b\beta + c} \right) \cr & = {a^2}{\beta ^2} + 2\left( {{a^2}{\beta ^2}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{Using eq}}{\text{.}}\left( 2 \right)} \right] \cr & = 3{a^2}{\beta ^2} > 0. \cr & {\text{Since }}\,f\left( \alpha \right)\,{\text{and }}f\left( \beta \right)\,{\text{are of opposite signs and }}\gamma {\text{ is a root}} \cr & {\text{of equation }}\,f\left( x \right) = 0 \cr & \therefore \,\,\gamma \,\,{\text{must lie between }}\alpha \,\,{\text{and}}\,\,\beta \cr & {\text{Thus }}\alpha < \gamma < \beta . \cr & \therefore \,\,\,\left( {\text{D}} \right){\text{is the correct option}}{\text{.}} \cr} $$

$$\alpha + \beta = \frac{{3a}}{{a - 1}},\alpha \beta = \frac{1}{{a - 1}}.\,{\text{As }}a > 1,\alpha + \beta > 0\,\,{\text{and }}\alpha \beta > 0.$$