Solve 4^2+5^2=x^2

$${ 4 }^{ 2 } + { 5 }^{ 2 } = { x }^{ 2 }$$

Solve for x

$$x=\sqrt{41}\approx 6.403124237$$

$$x=-\sqrt{41}\approx -6.403124237$$

Steps by Finding Square Root

Calculate 4 to the power of 2 and get 16.

$$16+5^{2}=x^{2}$$

Calculate 5 to the power of 2 and get 25.

$$16+25=x^{2}$$

Add 16 and 25 to get 41.

$$41=x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}=41$$

Take the square root of both sides of the equation.

$$x=\sqrt{41} x=-\sqrt{41}$$

Steps Using the Quadratic Formula

Calculate 4 to the power of 2 and get 16.

$$16+5^{2}=x^{2}$$

Calculate 5 to the power of 2 and get 25.

$$16+25=x^{2}$$

Add 16 and 25 to get 41.

$$41=x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}=41$$

Subtract 41 from both sides.

$$x^{2}-41=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -41 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-41\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-41\right)}}{2}$$

Multiply -4 times -41.

$$x=\frac{0±\sqrt{164}}{2}$$

Take the square root of 164.

$$x=\frac{0±2\sqrt{41}}{2}$$

Now solve the equation x=\frac{0±2\sqrt{41}}{2} when ± is plus.

$$x=\sqrt{41}$$

Now solve the equation x=\frac{0±2\sqrt{41}}{2} when ± is minus.

$$x=-\sqrt{41}$$

The equation is now solved.

$$x=\sqrt{41} x=-\sqrt{41}$$