Quadratic Equation MCQ Questions & Answers in Algebra | Maths

Can hold if $$x + y = 2x - y, x - 2y = 0, x - y - 1 = 0.$$

The roots of the given equation are
$$x = \frac{{2\cos \theta \pm \sqrt {4{{\cos }^2}\theta - 4} }}{2} = \cos \theta \pm i\sin \theta $$
Let, $$\alpha = \cos \theta + i\sin \theta \,\& \,\beta = \cos \theta - i\sin \theta $$
Then, $${\alpha ^n} = \cos n\theta + i\sin n\theta $$
$${\beta ^n} = \cos n\theta - i\sin n\theta $$
[Using De Moivre Theorem]
$${\alpha ^n} + {\beta ^n} = 2\cos n\theta \,\,{\text{and }}{\alpha ^n} \cdot {\beta ^n} = 1$$
∴ The required equation is
$${x^2} - 2x\cos n\theta + 1 = 0$$

$$\eqalign{ & {\text{As }}\left( {{x^2} + px + 1} \right){\text{ is a factor of }}a{x^3} + bx + c, \cr & {\text{we can assume that zeros of }}{x^2} + px + 1{\text{ are }}\alpha ,\beta {\text{ and that of}} \cr & a{x^3} + bx + c{\text{ be }}\alpha ,\beta ,\gamma {\text{ so that}} \cr & \alpha + \beta = - p\,\,\,\,\,\,\,\,\,.....\left( {\text{i}} \right) \cr & \alpha \beta = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.....\left( {{\text{ii}}} \right) \cr & {\text{and }}\alpha + \beta + \gamma = 0\,\,\,\,.....\left( {{\text{iii}}} \right) \cr & \alpha \beta + \beta \gamma + \gamma \alpha = \frac{b}{a}\,\,\,\,.....\left( {{\text{iv}}} \right) \cr & \alpha \beta \gamma = - \frac{c}{a}\,\,\,\,\,\,\,\,\,\,.....\left( {\text{v}} \right) \cr & {\text{Solving }}\left( {{\text{ii}}} \right){\text{and}}\left( {\text{v}} \right){\text{ we get }}\gamma = - \frac{c}{a}. \cr & {\text{Also from }}\left( {\text{i}} \right){\text{and}}\left( {{\text{iii}}} \right){\text{ we get }}\gamma = p \cr & \therefore \,\,\,p = \gamma = - \frac{c}{a} \cr & \,\,{\text{Using equations }}\left( {\text{i}} \right),\left( {{\text{ii}}} \right){\text{and}}\left( {{\text{iv}}} \right){\text{ we get}} \cr & \,\,\,\,1 + \gamma \left( { - p} \right) = \frac{b}{a} \cr & \Rightarrow \,\,1 + \left( { - \frac{c}{a}} \right)\left( {\frac{c}{a}} \right) = \frac{b}{a}\,\,\,\,\,\,\,\,\,\,\left( {{\text{using }}\gamma {\text{ = }}p = - \frac{c}{a}} \right) \cr & \Rightarrow \,\,1 - \frac{{{c^2}}}{{{a^2}}} = \frac{b}{a} \cr & \Rightarrow \,\,{a^2} - {c^2} = ab \cr & \therefore \,\,\,\left( {\text{C}} \right){\text{is the correct answer}}{\text{.}} \cr} $$

Given equation is
$$\left( {l - m} \right){x^2} + lx + 1 = 0$$
Roots are $$\alpha ,\beta .$$
∵ One root is double the other.
$$\beta = 2\alpha $$
Sum of roots $$ = \alpha + \beta .$$
$$\eqalign{ & 3\alpha = \frac{{ - l}}{{l - m}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha \left( {2\alpha } \right) = \frac{1}{{\left( {l - m} \right)}} \cr & \Rightarrow {\alpha ^2} = \frac{{{l^2}}}{{9{{\left( {l - m} \right)}^2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,2{\alpha ^2} = \frac{1}{{l - m}} \cr & \Rightarrow 2\frac{{{l^2}}}{{9{{\left( {l - m} \right)}^2}}} = \frac{1}{{\left( {l - m} \right)}} \cr & \Rightarrow \frac{{2{l^2}}}{{9\left( {l - m} \right)}} = 1 \cr & \Rightarrow 2{l^2} = 9\left( {l - m} \right) \cr & \Rightarrow 2{l^2} - 9l + 9m = 0 \cr} $$
For $$l$$ to be real discriminant should be $${b^2} - 4ac \geqslant 0$$
$$\eqalign{ & 81 - 4 \times 2 \times 9m \geqslant 0 \cr & m \leqslant \frac{9}{8}. \cr} $$

Roots of the second equation are reciprocal of those of the first.
$$\therefore \,\,{c_1}{x^2} + {b_1}x + {a_1} = 0\,\,{\text{and }}{a_2}{x^2} + {b_2}x + {c_2} = 0$$          have both roots common.

$$\eqalign{ & {\text{We have,}} \cr & f\left( x \right) = a{\left( {x + 1} \right)^2} + b\left( {{x^2} - 3x - 2} \right) + x + 1 \cr & {\text{On expanding }}{\left( {x + 1} \right)^2}{\text{,}}\,{\text{we get}} \cr & f\left( x \right) = a\left( {{x^2} + 2x + 1} \right) + b\left( {{x^2} - 3x - 2} \right) + x + 1 = 0 \cr & f\left( x \right) = {x^2}\left( {a + b} \right) + x\left( {2a - 3b + 1} \right) + \left( {a - 2b + 1} \right) \cr & {\text{If the above quadratic equation }}f\left( x \right) = 0\forall x = 0,{\text{ then}} \cr & a + b = 0......\left( 1 \right) \cr & 2a - 3b + 1 = 0......\left( 2 \right) \cr & a - 2b + 1 = 0......\left( 3 \right) \cr & {\text{From equation }}\left( 1 \right),\,a = - b \cr & {\text{Put }}a = - b{\text{ in equation }}\left( 2 \right){\text{ and }}\left( 3 \right),{\text{ we get}} \cr & b = \frac{1}{5}{\text{ from equation }}\left( 2 \right){\text{ and}} \cr & b = \frac{1}{3}{\text{ from equation }}\left( 3 \right),{\text{ which is not possible}}{\text{.}} \cr & \therefore \,\left( {a,\,b} \right) \in \phi {\text{ for which }}f\left( x \right) = 0\forall x \in R \cr} $$

Given equation can be written as
$$\eqalign{ & \left( {6k + 2} \right){x^2} + rx + 3k - 1 = 0\,\,\,\,\,.....\left( {\text{i}} \right) \cr & {\text{and, }}2\left( {6k + 2} \right){x^2} + px + 2\left( {3k - 1} \right) = 0\,\,\,\,\,.....\left( {{\text{ii}}} \right) \cr} $$
Condition for common root is
$$\frac{{12k + 4}}{{6k + 2}} = \frac{p}{r} = \frac{{6k - 2}}{{3k - 1}} = 2\,\,\,{\text{or, }}2r - p = 0$$

$$\eqalign{ & p + q = - p\,\,{\text{and }}\,pq = q \cr & \Rightarrow \,\,q\left( {p - 1} \right) = 0 \cr & \Rightarrow \,\,q = 0\,\,{\text{or }}p = 1. \cr & {\text{If }}\,q = 0,\,{\text{then }}p = 0.\,\,{\text{i}}{\text{.e}}{\text{.}}\,p = q \cr & \therefore \,\,p = 1\,\,{\text{and }}q = - 2. \cr} $$

$$\eqalign{ & {x^2} - 3x < 4 \cr & \Rightarrow {x^2} - 3x - 4 < 0 \cr & \Rightarrow {x^2} - 4x + x - 4 < 0 \cr & \Rightarrow x\left( {x - 4} \right) + 1\left( {x - 4} \right) < 0 \cr & \Rightarrow \left( {x + 1} \right)\left( {x - 4} \right) < 0 \cr & \Rightarrow x \in \left( { - 1,\,4} \right) \cr & {\text{Hence integer point are }}0,\,1,\,2,\,3 \cr} $$

Let $$\alpha ,\beta $$  be the roots of $$a{x^2} - bx - c = 0$$    and let $$\alpha ',\beta '$$  be the roots of $$a'{x^2} - b'x - c' = 0$$    such that $$\left| {\alpha - \beta } \right| = \left| {\alpha ' - \beta '} \right|,\,{\text{i}}{\text{.e}}{\text{.,}}\,{\left( {\alpha + \beta } \right)^2} - 4\alpha \beta = {\left( {\alpha ' + \beta '} \right)^2} - 4\alpha '\beta '$$
$$ \Rightarrow \,\,\frac{{{b^2} + 4ac}}{{{a^2}}} = \frac{{b{'^2} + 4a'c'}}{{a{'^2}}}.$$
Hence, the expression $$\frac{{{b^2} + 4ac}}{{{a^2}}}$$   does not vary in value.