Solve x^2=sqrt{25}+sqrt{36}

$${ x }^{ 2 } = \sqrt{ 25 } + \sqrt{ 36 }$$

Solve for x

$$x=\sqrt{11}\approx 3.31662479$$

$$x=-\sqrt{11}\approx -3.31662479$$

Steps by Finding Square Root

Calculate the square root of 25 and get 5.

$$x^{2}=5+\sqrt{36}$$

Calculate the square root of 36 and get 6.

$$x^{2}=5+6$$

Add 5 and 6 to get 11.

$$x^{2}=11$$

Take the square root of both sides of the equation.

$$x=\sqrt{11} x=-\sqrt{11}$$

Steps Using the Quadratic Formula

Calculate the square root of 25 and get 5.

$$x^{2}=5+\sqrt{36}$$

Calculate the square root of 36 and get 6.

$$x^{2}=5+6$$

Add 5 and 6 to get 11.

$$x^{2}=11$$

Subtract 11 from both sides.

$$x^{2}-11=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-11\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-11\right)}}{2}$$

Multiply -4 times -11.

$$x=\frac{0±\sqrt{44}}{2}$$

Take the square root of 44.

$$x=\frac{0±2\sqrt{11}}{2}$$

Now solve the equation x=\frac{0±2\sqrt{11}}{2} when ± is plus.

$$x=\sqrt{11}$$

Now solve the equation x=\frac{0±2\sqrt{11}}{2} when ± is minus.

$$x=-\sqrt{11}$$

The equation is now solved.

$$x=\sqrt{11} x=-\sqrt{11}$$