Quadratic Equation MCQ Questions & Answers in Algebra | Maths

$$a{x^2} + bx + c = 0,\,\,\alpha + \beta = \frac{{ - b}}{a},\alpha \beta = \frac{c}{a}$$
As for given condition, $$\alpha + \beta = \frac{1}{{{\alpha ^2}}} + \frac{1}{{{\beta ^2}}}$$
$$\eqalign{ & \alpha + \beta = \frac{{{\alpha ^2} + {\beta ^2}}}{{{\alpha ^2}{\beta ^2}}} - \frac{b}{a} = \frac{{\frac{{{b^2}}}{{{a^2}}} - \frac{{2c}}{a}}}{{\frac{{{c^2}}}{{{a^2}}}}} \cr & {\text{On simplification }}2{a^2}c = a{b^2} + b{c^2} \cr & \Rightarrow \,\,\frac{{2a}}{b} = \frac{c}{a} + \frac{b}{c} \cr & \Rightarrow \,\,\frac{c}{a},\frac{a}{b},\frac{b}{c}\,\,{\text{are in A}}{\text{.P}}{\text{.}} \cr & \therefore \,\,\frac{a}{c},\frac{b}{a}\& \frac{c}{b}\,\,{\text{are in H}}{\text{.P}}{\text{.}}\,\, \cr} $$

$$\eqalign{ & {\left( {{2^x}} \right)^2} - \left( {a - 4} \right){2^x} - {2^x} + \left( {a - 4} \right) = 0 \cr & \Rightarrow \,\,\left( {{2^x} - a + 4} \right)\left( {{2^x} - 1} \right) = 0 \cr & \therefore \,\,{2^x} = 1,a - 4.\,\,{\text{As }}x \leqslant 0,0 < a - 4 \leqslant 1. \cr} $$

$${x^{{3^n}}} + {\left( { - x} \right)^{{3^n}}} = 0\,\,{\text{if }}n \geqslant 0$$      and $$n$$ is an integer.

$$\eqalign{ & {\text{If }}x = n \in Z,{n^2} + {n^2} > 25.\,{\text{So, }}{n^2} > \frac{{25}}{2} \cr & \therefore \,\,x = n = 4,5,6,.....\,\,{\text{or, }} - 4, - 5, - 6,..... \cr & {\text{If }}x = n + k,n \in Z,0 < k < 1\,\,{\text{then }}{n^2} + {\left( {n + 1} \right)^2} > 25 \cr & {\text{or, }}{n^2} + n - 12 > 0 \cr & \therefore \,\,n < - 4\,\,{\text{or, }}n > 3 \cr & \therefore \,\,x < - 4 + k\,\,{\text{or, }}x > 3 + k,\,{\text{where }}0 < k < 1 \cr & \therefore \,\,x \leqslant - 4\,\,{\text{or, }}x \geqslant 4. \cr} $$

$${x^2} + px + q = 0\,$$
Sum of roots $$ = \,\tan {30^ \circ } + \tan {15^ \circ } = - p$$
Product of roots $$ = \,\tan {30^ \circ }.\tan {15^ \circ } = q$$
$$\eqalign{ & \tan {45^ \circ } = \frac{{\tan {{30}^ \circ } + \tan {{15}^ \circ }}}{{1 - \tan {{30}^ \circ }.\tan {{15}^ \circ }}} = \frac{{ - p}}{{1 - q}} = 1 \cr & \Rightarrow \,\, - p = 1 - q \cr & \Rightarrow \,\,\,q - p = 1 \cr & \therefore \,\,2 + q - p = 3 \cr} $$

$$\eqalign{ & \frac{{{x^2} - 3x + 4}}{{x + 1}} > 1 \cr & \Rightarrow \,\,\frac{{{x^2} - 4x + 3}}{{x + 1}} > 0 \cr & \Rightarrow \,\,\left( {x - 1} \right)\left( {x - 3} \right)\left( {x + 1} \right) > 0,x \ne - 1\left\{ {{\text{multiplying by }}{{\left( {x + 1} \right)}^2}} \right\} \cr} $$
∴ from general sign scheme:

$$\left\{ {\because \,\,{\text{for }}x = 0,\,{\text{expression}} > 0} \right\}.$$

As A.M. $$ \geqslant $$ G,M, for positive real numbers, we get
$$\eqalign{ & \frac{{\left( {a + b} \right) + \left( {c + d} \right)}}{2} \geqslant \sqrt {\left( {a + b} \right)\left( {c + d} \right)} \cr & \Rightarrow \,M \leqslant 1 \left( {{\text{Putting values}}} \right) \cr & {\text{Also }}\left( {a + b} \right)\left( {c + d} \right) > 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\therefore \,\,a,b,c,d > 0} \right] \cr & \therefore \,\,0 \leqslant M \leqslant 1 \cr} $$

Let $$xy + yz + zx = \lambda .$$    Then
$$\eqalign{ & {x^2} + {y^2} + {z^2} - \lambda = \frac{1}{2}\left[ {{{\left( {x - y} \right)}^2} + {{\left( {y - z} \right)}^2} + {{\left( {z - x} \right)}^2}} \right] \geqslant 0 \cr & \Rightarrow \,\,1 - \lambda \geqslant 0. \cr & {\text{Again, }}{\left( {x + y + z} \right)^2} = {x^2} + {y^2} + {z^2} + 2\lambda = 1 + 2\lambda \cr & \Rightarrow \,\,1 + 2\lambda \geqslant 0. \cr} $$

Given equation is
$$\eqalign{ & {x^2} + \left( {2 + \lambda } \right)x - \frac{1}{2}\left( {1 + \lambda } \right) = 0 \cr & {\text{So, }}\alpha + \beta = - \left( {2 + \lambda } \right) = 0{\text{ and }}\alpha \beta = - \left( {\frac{{1 + \lambda }}{2}} \right) \cr & {\text{Now, }}{\alpha ^2} + {\beta ^2} = {\left( {\alpha + \beta } \right)^2} - 2\alpha \beta \cr & \Rightarrow {\alpha ^2} + {\beta ^2} = {\left[ { - \left( {2 + \lambda } \right)} \right]^2} + 2\frac{{\left( {1 + \lambda } \right)}}{2} \cr & \Rightarrow {\alpha ^2} + {\beta ^2} = {\lambda ^2} + 4 + 4\lambda + 1 + \lambda \cr & = {\lambda ^2} + 5\lambda + 5 \cr} $$
Which is minimum for $$\lambda = \frac{1}{2}.$$

Let the roots of equation $${x^2} - 6x + a = 0\,$$   be $$\alpha $$ and $$4\beta $$  and that of the equation
$${x^2} - cx + 6 = 0$$    be $$\alpha $$ and $$3\beta $$ . Then
$$\eqalign{ & \,\,\,\,\,\,\,\,\,\alpha + \,4\beta = 6;\,\,\,\,\,\,\,4\alpha \beta = a \cr & {\text{and }}\alpha {\text{ + 3}}\beta = c;\,\,\,\,\,\,\,3\alpha \beta = 6 \cr & \Rightarrow \,\,a = 8 \cr & \therefore \,\,{\text{The equation becomes }}\,{x^2} - 6x + 8 = 0 \cr & \Rightarrow \,\,\left( {x - 2} \right)\left( {x - 4} \right) = 0 \cr & \Rightarrow \,\,{\text{roots are 2 and 4}} \cr & \Rightarrow \,\,\alpha = 2,\beta = 1 \cr & \therefore \,\,{\text{Common root is 2}}{\text{.}} \cr} $$