Quadratic Equation MCQ Questions & Answers in Algebra | Maths

$$6{x^2} - xy - {y^2} - 6x + 8y - 12 = 6\left( {x + \lambda y - 2} \right)\left( {x - \mu y + 1} \right).$$
Equate co-efficients and solve for $$\lambda ,\mu .$$

As the co-efficients are in reverse order, the roots of $$a{x^2} + bx + c = 0$$    are $$\alpha ,\beta ,$$  while the roots of $$c{x^2} + bx + a = 0$$    are $$\frac{1}{\alpha },\frac{1}{\beta }.$$
One negative root is common
$$\eqalign{ & \Rightarrow \,\,\left( {\text{i}} \right)\alpha = \frac{1}{\alpha } < 0;\,{\text{so, }}\alpha = - 1 \cr & {\text{or, }}\left( {{\text{ii}}} \right)\alpha = \frac{1}{\beta } < 0 \cr & \Rightarrow \,\,\alpha \beta = 1 \cr & \Rightarrow \,\,\frac{c}{a} = 1 \cr & \Rightarrow \,\,c = a\left( {{\text{not possible}}} \right). \cr} $$

$$D = {c^2}{\left( {3{a^2} + {b^2}} \right)^2} - 4ab{c^2}\left( { - 6{a^2} - ab + 2{b^2}} \right) = {c^2}{\left( {3{a^2} - {b^2} + 4ab} \right)^2}.$$

$$\eqalign{ & \alpha - p > 0,\beta - p > 0\,\,{\text{and }}D \geqslant 0 \cr & \Rightarrow \,\,\alpha + \beta - 2p > 0,\alpha \beta - p\left( {\alpha + \beta } \right) + {p^2} > 0\,\,{\text{and }}{p^2} + 4{p^2} \geqslant 0 \cr & \therefore \,\,p - 2p > 0, - {p^2} - p \cdot p + {p^2} > 0 \cr & \Rightarrow \,\,p < 0,{p^2} < 0\left( {{\text{absurd}}} \right). \cr} $$

$$\eqalign{ & {x^2} + 6x - 7 < 0 \cr & \Rightarrow \left( {x + 7} \right)\left( {x - 1} \right) < 0 \cr & \Rightarrow x = \left( { - 7,1} \right) \cr & \Rightarrow A = \left\{ { - 7, - 6, - 5, - 4, - 3, - 2, - 1,0,1} \right\} \cr & \Rightarrow {x^2} + 9x + 14 > 0 \cr & \Rightarrow \left( {x + 7} \right)\left( {x + 2} \right) > 0 \cr & \Rightarrow x = \left( { - \infty , - 7} \right) \cup \left( { - 2,\infty } \right) \cr & \Rightarrow B = R - \left\{ { - 7, - 6, - 5, - 4, - 3, - 2} \right\} \cr} $$

$$\eqalign{ & {\text{So, }}A \cap B = \left( { - 2,1} \right) \cr & \frac{A}{B} = \left( { - 7, - 2} \right). \cr} $$

The given equations are
$$\eqalign{ & x + 2y + 2z = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.....\left( 1 \right) \cr & {\text{and }}2x + 4y + 4z = 9\,\,\,\,.....\left( 2 \right) \cr & {\text{Subtracting }}\left( 1 \right) \times \left( 2 \right){\text{from}}\left( 2 \right),{\text{ we get 0}} = {\text{7}} \cr & \left( {{\text{not possible}}} \right) \cr & \therefore {\text{ No solution}}{\text{.}} \cr} $$

$${\sin ^2}\frac{x}{2} \cdot {\cos ^2}x = \frac{{x + \frac{1}{x}}}{2} \geqslant \sqrt {x \cdot \frac{1}{x}} = 1,$$        equality will hold when $$x = \frac{1}{x}.$$
$$\therefore \,\,{\sin^2}\frac{x}{2} \cdot {\cos ^2}x = 1\,\,{\text{and then }}x = \frac{1}{x},\,{\text{i}}{\text{.e}}{\text{., }}x = 1$$          for which the equation is not satisfied.

Clearly, there are two points of intersection of $$y = \left| x \right|\,\,{\text{and }}y = \cos x.$$
Hence, there are two real solutions.

$$\eqalign{ & {x^2} - 3\left| x \right| + 2 = 0\, \cr & \Rightarrow \,\,{\left| x \right|^2} - 3\left| x \right| + 2 = 0 \cr & \left( {\left| x \right| - 2} \right)\left( {\left| x \right| - 1} \right) = 0 \cr & \left| x \right| = 1,2\,\,{\text{or }}x = \pm 1, \pm 2 \cr & \therefore \,\,{\text{No of solution}} = 4 \cr} $$

Let $$\alpha ,\beta \,\,{\text{and }}\gamma {\text{,}}\delta $$   be the roots of the equations
$${x^2} + ax + b = 0\,\,{\text{and }}\,{x^2} + bx + a = 0$$        respectively.
$$\eqalign{ & \therefore \,\,\alpha + \beta = - a,\,\alpha \beta = b\,\,{\text{and }}\gamma {\text{ + }}\delta = - b,\,\gamma \delta = a. \cr & {\text{Given }}\left| {\alpha - \beta } \right| = \left| {\gamma - \delta } \right| \cr & \Rightarrow \,\,{\left( {\alpha - \beta } \right)^2} = {\left( {\gamma - \delta } \right)^2} \cr & \Rightarrow \,\,{\left( {\alpha + \beta } \right)^2} - 4\alpha \beta = {\left( {\gamma + \delta } \right)^2} - 4\gamma \delta \cr & \Rightarrow \,\,{a^2} - 4b = {b^2} - 4a \cr & \Rightarrow \,\,\left( {{a^2} - {b^2}} \right) + 4\left( {a - b} \right) = 0 \cr & \Rightarrow \,\,a + b + 4 = 0\,\,\,\,\,\,\,\,\,\,\,\,\left( {\because \,\,a \ne b} \right) \cr} $$