The sum of the squares of the roots of |x-2|^ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) For the equation $|x-2|^2+|x-2|-2=0$:Let $t = |x-2|$,then $t^2+t-2=0$.$(t+2)(t-1)=0 \Rightarrow t=1$ (since $t \geq 0$).$|x-2|=1 \Rightarrow x-2=1

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(B) For the equation $|x-2|^2+|x-2|-2=0$:Let $t = |x-2|$,then $t^2+t-2=0$.$(t+2)(t-1)=0 \Rightarrow t=1$ (since $t \geq 0$).$|x-2|=1 \Rightarrow x-2=1$ or $x-2=-1$.$x=3, 1$.Sum of squares of roots $= 3^2+1^2 = 9+1 = 10$.For the equation $x^2-2|x-3|-5=0$:Case-$I$: $x \geq 3$,then $x^2-2(x-3)-5=0$ $\Rightarrow x^2-2x+1=0$ $\Rightarrow (x-1)^2=0$ $\Rightarrow x=1$.Since $1 Case-$II$: $x Roots $\alpha, \beta = \frac{-2 \pm \sqrt{4 - 4(1)(-11)}}{2} = -1 \pm 2\sqrt{3}$.Both roots are less than $3$,so both are valid.Sum of squares of roots $= (\alpha+\beta)^2 - 2\alpha\beta = (-2)^2 - 2(-11) = 4 + 22 = 26$.Total sum $= 10 + 26 = 36$.

The sum of the squares of the roots of $|x-2|^2+|x-2|-2=0$ and the squares of the roots of $x^2-2|x-3|-5=0$ is:

Similar Questions

If $x$ is real,the function $f(x) = \frac{(x - a)(x - b)}{(x - c)}$ will assume all real values,provided

Let $x, y, z$ be positive real numbers. Which of the following conditions imply $x=y=z$?
$I.$ $x^3+y^3+z^3=3xyz$
$II.$ $x^3+y^2z+yz^2=3xyz$
$III.$ $x^3+y^2z+z^2x=3xyz$
$IV.$ $(x+y+z)^3=27xyz$

The set of all real values of the expression $\frac{x^2-x+2}{x^2+x-2}$ for all $x \in R-\{-2, 1\}$ is

The values of $\frac{x^2-2x+1}{x^2+x-1}$ do not lie in the interval

Let $\alpha, \beta$ be the roots of the equation $x^{2}-4 \lambda x+5=0$ and $\alpha, \gamma$ be the roots of the equation $x^{2}-(3 \sqrt{2}+2 \sqrt{3}) x+7+3 \lambda \sqrt{3}=0$. If $\beta+\gamma=3 \sqrt{2}$,then $(\alpha+2 \beta+\gamma)^{2}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module