Question

$\alpha ,\beta$ are roots of the equation $2x^2-5x-6=0$ then

• A. the equation whose roots are $\alpha ,\beta$ and 2 is $2x^3-9x^2+4x+12=0$
• B. The value of $\left(\frac{\alpha }{\beta }+\frac{\beta }{\alpha }+1\right)$ is $\frac{-37}{12}$
• C. the equation whose roots are $\alpha ,\beta$ and -1 is $2x^3-3x^2-11x-6=0$
• D. The value of $(\alpha -\beta {)}^2$ is $\frac{53}{4}$

Answer 1

The correct option is C the equation whose roots are $\alpha ,\beta$ and -1 is $2x^3-3x^2-11x-6=0$

${2x}^2-5x-6=0$

has roots $\alpha ,\beta$

So, equation having roots $\alpha ,\beta ,2$ will be

$({2x}^2-5x-6)(x-2)=0\Rightarrow {2x}^3-{9x}^2+4x+12=0$

$\frac{\alpha }{\beta }+\frac{\beta }{\alpha }+1=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }+1=\frac{{(\alpha +\beta )}^2}{\alpha \beta }+1=\frac{-49}{12}$

as $\alpha +\beta =5/2$ and $\alpha \beta =\frac{-6}{2}=-3$

${(\alpha -\beta )}^2={\alpha }^2+{\beta }^2-2\alpha \beta ={(\alpha -\beta )}^2-4\alpha \beta =\frac{73}{4}$

Equation whose roots are $\alpha ,\beta ,-1$ will be

$({2x}^2-5x-6)(x+1)=0\Rightarrow {2x}^3-{3x}^2-11x-6=0$