Quadratic Equation MCQ Questions & Answers in Algebra | Maths

$$\because \,\,\alpha ,\beta \,\,$$ are the roots of $${x^2} - 6x - 2 = 0$$
$$\eqalign{ & \therefore \,\,\,{\alpha ^2} - 6\alpha - 2 = 0 \cr & \Rightarrow \,\,{\alpha ^{10}} - 6{\alpha ^9} - 2{\alpha ^8} = 0 \cr & \Rightarrow \,\,{\alpha ^{10}} - 2{\alpha ^8} = 6{\alpha ^9}\,\,\,\,\,\,\,\,\,\,\,\,.....\left( 1 \right) \cr & {\text{Similarly }}{\beta ^{10}} - 2{\beta ^8} = 6{\beta ^9}\,\,\,\,\,\,.....\left( 2 \right) \cr} $$
From equation (1) and (2)
$$\eqalign{ & {\alpha ^{10}} - {\beta ^{10}} - 2\left( {{\alpha ^8} - {\beta ^8}} \right) = 6\left( {{\alpha ^9} - {\beta ^9}} \right) \cr & \Rightarrow \,\,{a_{10}} - 2{a_8} = 6{a_9} \cr & \Rightarrow \,\,\frac{{{a_{10}} - 2{a_8}}}{{2{a_9}}} = 3 \cr} $$

Let $$f\left( x \right) = {a^2}{x^2} + 2bx + 2c.$$     From the question, $${a^2}{\alpha ^2} + b\alpha + c = 0$$    and $${a^2}{\beta ^2} - b\beta - c = 0.$$
Now, $$f\left( \alpha \right) = {a^2}{\alpha ^2} + 2b\alpha + 2c = b\alpha + c = - {a^2}{\alpha ^2}$$
$$f\left( \beta \right) = {a^2}{\beta ^2} + 2b\beta + 2c = 3b\beta + 3c = 3\left( {b\beta + c} \right) = 3{a^2}{\beta ^2}.$$
Clearly, $$0 < \alpha < \beta $$
⇒ $$\alpha ,\beta $$  are real. So, $$f\left( \alpha \right) < 0,f\left( \beta \right) > 0.$$
So, $$f\left( \gamma \right) = 0$$   where $$\alpha < \gamma < \beta .$$

$$\eqalign{ & y = \frac{{3{x^2} + 9x + 17}}{{3{x^2} + 9x + 7}} \cr & 3{x^2}\left( {y - 1} \right) + 9x\left( {y - 1} \right) + 7y - 17 = 0 \cr & D \geqslant 0\,\,\,\,\,\,\,\,\,\because \,\,x\,\,{\text{is real}} \cr & {\text{81}}{\left( {y - 1} \right)^2} - 4 \times 3\left( {y - 1} \right)\left( {7y - 17} \right) \geqslant 0 \cr & \Rightarrow \,\,\left( {y - 1} \right)\left( {y - 41} \right) \leqslant 0 \cr & \Rightarrow \,\,1 \leqslant y \leqslant 41 \cr & \therefore \,\,{\text{Max value of }}y{\text{ is 41}}{\text{.}} \cr} $$

As it is a third-degree homogeneous expression in $$x, y,$$  we have $${y^3} + {y^2}x - \mu y{x^2} + \lambda {x^3}$$
$$\eqalign{ & = \left( {y + x} \right)\left( {y + 3x} \right)\left( {y + mx} \right) \cr & = {y^3} + \left( {m + 3 + 1} \right){y^2}x + \left( {3 + m + 3m} \right)y{x^2} + 3m{x^3} \cr & \Rightarrow \,\,1 = m + 4, - \mu = 3 + 4m,\lambda = 3m. \cr} $$

$$a\left( {1 - 2{{\sin }^2}x} \right) + b{\sin ^2}x + c = 0,\,{\text{i}}{\text{.e}}{\text{., }}\left( {b - 2a} \right){\sin ^2}x + \left( {a + c} \right) = 0.$$
It is an identity if $$b - 2a = 0,a + c = 0.\,{\text{So, }}\frac{a}{1} = \frac{b}{2} = \frac{c}{{ - 1}}.$$