The sum of the solutions of the equation |sqr | vedclass

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

Question

(C) Given equation: $|\sqrt{x} - 2| + \sqrt{x}(\sqrt{x} - 4) + 2 = 0$.Let $t = \sqrt{x}$,where $t > 0$. The equation becomes $|t - 2| + t(t - 4) + 2 =

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(C) Given equation: $|\sqrt{x} - 2| + \sqrt{x}(\sqrt{x} - 4) + 2 = 0$.Let $t = \sqrt{x}$,where $t > 0$. The equation becomes $|t - 2| + t(t - 4) + 2 = 0$.$|t - 2| + t^2 - 4t + 2 = 0$.We can rewrite $t^2 - 4t + 2$ as $(t^2 - 4t + 4) - 2 = (t - 2)^2 - 2$.So,$|t - 2| + (t - 2)^2 - 2 = 0$.Let $u = |t - 2|$. Since $u^2 = |t - 2|^2 = (t - 2)^2$,the equation becomes $u + u^2 - 2 = 0$.$u^2 + u - 2 = 0$.$(u + 2)(u - 1) = 0$.This gives $u = -2$ or $u = 1$.Since $u = |t - 2| \ge 0$,$u = -2$ is rejected.Thus,$|t - 2| = 1$.This implies $t - 2 = 1$ or $t - 2 = -1$.$t = 3$ or $t = 1$.Since $t = \sqrt{x}$,we have $\sqrt{x} = 3 \implies x = 9$ and $\sqrt{x} = 1 \implies x = 1$.Both values satisfy $x > 0$.The sum of the solutions is $9 + 1 = 10$.

The sum of the solutions of the equation $|\sqrt{x} - 2| + \sqrt{x}(\sqrt{x} - 4) + 2 = 0$ $(x > 0)$ is equal to

Similar Questions

If $\alpha$ and $\beta$ are the roots of the quadratic equation $ax^{2} + bx + c = 0$,the value of $\alpha^{3} + \beta^{3}$ is

If a root of the given equation $a(b - c)x^2 + b(c - a)x + c(a - b) = 0$ is $1$,then the other root will be

Let $\alpha$ and $\beta$ be the roots of the equation $x^2 + x + 1 = 0$. The equation whose roots are $\alpha^{19}$ and $\beta^7$ is:

If $x = \sqrt{7 + 4\sqrt{3}},$ then $x + \frac{1}{x} = $

Solve the given two equations and select the correct option.
$I.$ $x^{2}-5x+6=0$
$II.$ $2y^{2}-15y+27=0$

Vedclass Test Series

Exam Paper Generator

Online Exam Module