The solution to the equation sqrt{x + 10} + s | vedclass

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

Question

(B) Given equation: $\sqrt{x + 10} + \sqrt{x - 2} = 6$ Let $u = \sqrt{x + 10}$ and $v = \sqrt{x - 2}$. Then $u^2 = x + 10$ and $v^2 = x - 2$. Subtract

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) Given equation: $\sqrt{x + 10} + \sqrt{x - 2} = 6$ Let $u = \sqrt{x + 10}$ and $v = \sqrt{x - 2}$. Then $u^2 = x + 10$ and $v^2 = x - 2$. Subtracting the two: $u^2 - v^2 = (x + 10) - (x - 2) = 12$. We know $(u - v)(u + v) = 12$. Since $u + v = 6$,we have $(u - v)(6) = 12$,which implies $u - v = 2$. Now we have a system: $u + v = 6$ $u - v = 2$ Adding these: $2u = 8 \implies u = 4$. Since $u = \sqrt{x + 10} = 4$,squaring both sides gives $x + 10 = 16$,so $x = 6$. Checking the value: $\sqrt{6 + 10} + \sqrt{6 - 2} = \sqrt{16} + \sqrt{4} = 4 + 2 = 6$. The solution is $x = 6$.

The solution to the equation $\sqrt{x + 10} + \sqrt{x - 2} = 6$ is:

Similar Questions

If the equation $x^4+ax^3+bx^2+cx+d=0$ has three equal roots,then that root is

If $-1+i$ is a root of the equation $x^4+4x^3+5x^2+2x-2=0$,then the real roots of this equation are

If $1-i$ is a root of the equation $x^2+ax+b=0$,then $b$ is equal to

The sum of all the roots of the equation $(x-1)^{2}-5|x-1|+6=0$ is:

If $\alpha$ satisfies the equation $\sqrt{\frac{x}{2x+1}} + \sqrt{\frac{2x+1}{x}} = 2$,then the roots of the equation $\alpha^2 x^2 + 4\alpha x + 3 = 0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module