Solve 25^2/75^2+x^2/45=1

$$\frac{ { 25 }^{ 2 } }{ { 75 }^{ 2 } } + \frac{ { x }^{ 2 } }{ 45 } =1$$

Solve for x

$$x=2\sqrt{10}\approx 6.32455532$$

$$x=-2\sqrt{10}\approx -6.32455532$$

Steps by Finding Square Root

Multiply both sides of the equation by 45.

$$\frac{1}{125}\times 25^{2}+x^{2}=45$$

Calculate 25 to the power of 2 and get 625.

$$\frac{1}{125}\times 625+x^{2}=45$$

Multiply \frac{1}{125} and 625 to get 5.

$$5+x^{2}=45$$

Subtract 5 from both sides.

$$x^{2}=45-5$$

Subtract 5 from 45 to get 40.

$$x^{2}=40$$

Take the square root of both sides of the equation.

$$x=2\sqrt{10} x=-2\sqrt{10}$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 45.

$$\frac{1}{125}\times 25^{2}+x^{2}=45$$

Calculate 25 to the power of 2 and get 625.

$$\frac{1}{125}\times 625+x^{2}=45$$

Multiply \frac{1}{125} and 625 to get 5.

$$5+x^{2}=45$$

Subtract 45 from both sides.

$$5+x^{2}-45=0$$

Subtract 45 from 5 to get -40.

$$-40+x^{2}=0$$

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.

$$x^{2}-40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-40\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-40\right)}}{2}$$

Multiply -4 times -40.

$$x=\frac{0±\sqrt{160}}{2}$$

Take the square root of 160.

$$x=\frac{0±4\sqrt{10}}{2}$$

Now solve the equation x=\frac{0±4\sqrt{10}}{2} when ± is plus.

$$x=2\sqrt{10}$$

Now solve the equation x=\frac{0±4\sqrt{10}}{2} when ± is minus.

$$x=-2\sqrt{10}$$

The equation is now solved.

$$x=2\sqrt{10} x=-2\sqrt{10}$$