The solution of 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: $\sqrt{x + 10} + \sqrt{x - 2} = 6$Rearranging the terms: $\sqrt{x + 10} = 6 - \sqrt{x - 2}$Squaring both sides: $x + 10 = (6 - \sq

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) Given equation: $\sqrt{x + 10} + \sqrt{x - 2} = 6$Rearranging the terms: $\sqrt{x + 10} = 6 - \sqrt{x - 2}$Squaring both sides: $x + 10 = (6 - \sqrt{x - 2})^2$$x + 10 = 36 + (x - 2) - 12\sqrt{x - 2}$$x + 10 = 34 + x - 12\sqrt{x - 2}$Subtracting $x$ from both sides: $10 = 34 - 12\sqrt{x - 2}$$-24 = -12\sqrt{x - 2}$Dividing by $-12$: $2 = \sqrt{x - 2}$Squaring both sides again: $4 = x - 2$$x = 6$Verification: $\sqrt{6 + 10} + \sqrt{6 - 2} = \sqrt{16} + \sqrt{4} = 4 + 2 = 6$. This satisfies the equation.

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

Similar Questions

If $x$ is real,then the minimum value of $y = \frac{x^2-x+1}{x^2+x+1}$ is

The value of $2 + \frac{1}{2 + \frac{1}{2 + \dots \infty}}$ is

If $f(x) \in \mathbb{Q}[x]$ is a non-zero polynomial such that all its roots are irrational,then the degree of $f(x)$ is

The midpoint of the diagonal of a rectangle formed by $x^2+5x-6=0$ and $y^2-8y-20=0$ is

If $p$ and $q$ are prime numbers satisfying the condition $p^{2}-2q^{2}=1$,then the value of $p^{2}+2q^{2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module