Let [], [] be the roots of the equation [x^2-√2 x+√6=0] and [1^2+1, 1^2+1] be the roots of…

Question

Let $$\alpha$$, $$\beta$$ be the roots of the equation $$x^{2}-\sqrt{2} x+\sqrt{6}=0$$ and $$\frac{1}{\alpha^{2}}+1, \frac{1}{\beta^{2}}+1$$ be the roots of the equation $x^{2}+a x+b=0$. Then the roots of the equation $x^{2}-(a+b-2) x+(a+b+2)=0$ are :

Accepted Answer

$$\alpha  + \beta  = \sqrt 2 $$, $$\alpha \beta  = \sqrt 6 $$

$${1 \over {{\alpha ^2}}} + 1 + {1 \over {{\beta ^2}}} + 1 = 2 + {{{\alpha ^2} + {\beta ^2}} \over 6}$$

$$ = 2 + {{2 - 2\sqrt 6 } \over 6} =  - a$$

$$\left( {{1 \over {{\alpha ^2}}} + 1} \right)\left( {{1 \over {{\beta ^2}}} + 1} \right) = 1 + {1 \over {{\alpha ^2}}} + {1 \over {{\beta ^2}}} + {1 \over {{\alpha ^2}{\beta ^2}}}$$

$$ = {7 \over 6} + {{2 - 2\sqrt 6 } \over 6} = b$$

$$ \Rightarrow a + b = {{ - 5} \over 6}$$

So, equation is $${x^2} + {{17x} \over 6} + {7 \over 6} = 0$$

OR $6{x^2} + 17x + 7 = 0$

Both roots of equation are -ve and distinct

Solutions · 1