Question

If a,b are the roots of the equation $x^2+2x+4=0,$ then $\frac{1}{a^2}+\frac{1}{b^2}=$

• A. $-\frac{1}{2}$
• B. $-\frac{1}{4}$
• C. 32
• D. 34

Answer 1

The correct option is B $-\frac{1}{4}$

Given Eq. is $x^2+2x+4=\mathrm{0.....}\left(1\right)$

since $a,b$ are roots of EQ (1)

sum of roots are $a+b=-2$

Product of roots are $a-b=4$

consider ${(a+b)}^2=(a^2+2ab+b^2)$

${(-2)}^2=a^2+2.4+b^2$

$a^2+b^2=4-8$

$a^2+b^2=-4$

Consider $\frac{1}{a^2}+\frac{1}{b^2}=\frac{b^2+a^2}{a^2+b^2}=\frac{a^2+b^2}{{(a.b)}^2}$

$=-\frac{4}{16}=-\frac{1}{4}$

$\therefore$ $\frac{1}{a^2}+\frac{1}{b^2}=-\frac{1}{4}$