Sign in
Please select an account to continue using cracku.in
↓ →
The sum of the squares of all the roots of the equation $$x^{2} + |x + 4| + |x − 4| − 35 = 0$$ is
Given equation is,
$$x^{2} + |x + 4| + |x − 4| − 35 = 0$$
i.) When $$x>4$$,
$$x^2+x+4+x-4-35=0$$
or, $$x^2+2x-35=0$$
or, $$x^2+7x-5x-35=0$$
or, $$\left(x+7\right)\left(x-5\right)=0$$
so, $$x=5,-7$$
But $$x=-7$$ is not acceptable as $$x>4$$
So, only acceptable value is $$x=5$$
ii.) When $$-4\le x\le4$$
$$x^2+x+4-x+4-35=0$$
or, $$x^2-27=0$$
or, $$x=\pm\ \sqrt{\ 27}$$
But $$\sqrt{\ 27}=5.196$$
So, the obtained values of $$x$$ does not lie in this range. Hence, the case is rejected.
iii.) When $$x<-4$$,
$$x^2-x-4+4-x-35=0$$
or, $$x^2-2x-35=0$$
or, $$x^2-7x+5x-35=0$$
or, $$\left(x-7\right)\left(x+5\right)=0$$
or, $$x=7,-5$$
But $$x=7$$ is not acceptable as $$x<-4$$
So, only acceptable value is $$x=-5$$
So, the roots of the equation are $$x=5,-5$$
Sum of square of roots =$$\left(5\right)^2+\left(-5\right)^2=25+25=50$$
Create a FREE account and get: