Solve 5^2-x^2=3^2

$${ 5 }^{ 2 } - { x }^{ 2 } = { 3 }^{ 2 }$$

Solve for x

$$x=4$$

$$x=-4$$

Steps by Finding Square Root

Calculate 5 to the power of 2 and get 25.

$$25-x^{2}=3^{2}$$

Calculate 3 to the power of 2 and get 9.

$$25-x^{2}=9$$

Subtract 25 from both sides.

$$-x^{2}=9-25$$

Subtract 25 from 9 to get -16.

$$-x^{2}=-16$$

Divide both sides by -1.

$$x^{2}=\frac{-16}{-1}$$

Fraction \frac{-16}{-1} can be simplified to 16 by removing the negative sign from both the numerator and the denominator.

$$x^{2}=16$$

Take the square root of both sides of the equation.

$$x=4 x=-4$$

Steps Using the Quadratic Formula

Calculate 5 to the power of 2 and get 25.

$$25-x^{2}=3^{2}$$

Calculate 3 to the power of 2 and get 9.

$$25-x^{2}=9$$

Subtract 9 from both sides.

$$25-x^{2}-9=0$$

Subtract 9 from 25 to get 16.

$$16-x^{2}=0$$

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$-x^{2}+16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 0 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-1\right)\times 16}}{2\left(-1\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-1\right)\times 16}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{0±\sqrt{4\times 16}}{2\left(-1\right)}$$

Multiply 4 times 16.

$$x=\frac{0±\sqrt{64}}{2\left(-1\right)}$$

Take the square root of 64.

$$x=\frac{0±8}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{0±8}{-2}$$

Now solve the equation x=\frac{0±8}{-2} when ± is plus. Divide 8 by -2.

$$x=-4$$

Now solve the equation x=\frac{0±8}{-2} when ± is minus. Divide -8 by -2.

$$x=4$$

The equation is now solved.

$$x=-4 x=4$$