Solve x^2/25+(4/9x+4/3)^2/16=1

$$\frac { x ^ { 2 } } { 25 } + \frac { ( \frac { 4 } { 9 } x + \frac { 4 } { 3 } ) ^ { 2 } } { 16 } = 1$$

Solve for x

$$x = \frac{45 \sqrt{97} - 75}{106} \approx 3.473571708$$

$$x=\frac{-45\sqrt{97}-75}{106}\approx -4.888666048$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 400, the least common multiple of 25,16.

$$16x^{2}+25\left(\frac{4}{9}x+\frac{4}{3}\right)^{2}=400$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\frac{4}{9}x+\frac{4}{3}\right)^{2}.

$$16x^{2}+25\left(\frac{16}{81}x^{2}+\frac{32}{27}x+\frac{16}{9}\right)=400$$

Use the distributive property to multiply 25 by \frac{16}{81}x^{2}+\frac{32}{27}x+\frac{16}{9}.

$$16x^{2}+\frac{400}{81}x^{2}+\frac{800}{27}x+\frac{400}{9}=400$$

Combine 16x^{2} and \frac{400}{81}x^{2} to get \frac{1696}{81}x^{2}.

$$\frac{1696}{81}x^{2}+\frac{800}{27}x+\frac{400}{9}=400$$

Subtract 400 from both sides.

$$\frac{1696}{81}x^{2}+\frac{800}{27}x+\frac{400}{9}-400=0$$

Subtract 400 from \frac{400}{9} to get -\frac{3200}{9}.

$$\frac{1696}{81}x^{2}+\frac{800}{27}x-\frac{3200}{9}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1696}{81} for a, \frac{800}{27} for b, and -\frac{3200}{9} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{800}{27}±\sqrt{\left(\frac{800}{27}\right)^{2}-4\times \frac{1696}{81}\left(-\frac{3200}{9}\right)}}{2\times \frac{1696}{81}}$$

Square \frac{800}{27} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{800}{27}±\sqrt{\frac{640000}{729}-4\times \frac{1696}{81}\left(-\frac{3200}{9}\right)}}{2\times \frac{1696}{81}}$$

Multiply -4 times \frac{1696}{81}.

$$x=\frac{-\frac{800}{27}±\sqrt{\frac{640000}{729}-\frac{6784}{81}\left(-\frac{3200}{9}\right)}}{2\times \frac{1696}{81}}$$

Multiply -\frac{6784}{81} times -\frac{3200}{9} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{800}{27}±\sqrt{\frac{640000+21708800}{729}}}{2\times \frac{1696}{81}}$$

Add \frac{640000}{729} to \frac{21708800}{729} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{800}{27}±\sqrt{\frac{2483200}{81}}}{2\times \frac{1696}{81}}$$

Take the square root of \frac{2483200}{81}.

$$x=\frac{-\frac{800}{27}±\frac{160\sqrt{97}}{9}}{2\times \frac{1696}{81}}$$

Multiply 2 times \frac{1696}{81}.

$$x=\frac{-\frac{800}{27}±\frac{160\sqrt{97}}{9}}{\frac{3392}{81}}$$

Now solve the equation x=\frac{-\frac{800}{27}±\frac{160\sqrt{97}}{9}}{\frac{3392}{81}} when ± is plus. Add -\frac{800}{27} to \frac{160\sqrt{97}}{9}.

$$x=\frac{\frac{160\sqrt{97}}{9}-\frac{800}{27}}{\frac{3392}{81}}$$

Divide -\frac{800}{27}+\frac{160\sqrt{97}}{9} by \frac{3392}{81} by multiplying -\frac{800}{27}+\frac{160\sqrt{97}}{9} by the reciprocal of \frac{3392}{81}.

$$x=\frac{45\sqrt{97}-75}{106}$$

Now solve the equation x=\frac{-\frac{800}{27}±\frac{160\sqrt{97}}{9}}{\frac{3392}{81}} when ± is minus. Subtract \frac{160\sqrt{97}}{9} from -\frac{800}{27}.

$$x=\frac{-\frac{160\sqrt{97}}{9}-\frac{800}{27}}{\frac{3392}{81}}$$

Divide -\frac{800}{27}-\frac{160\sqrt{97}}{9} by \frac{3392}{81} by multiplying -\frac{800}{27}-\frac{160\sqrt{97}}{9} by the reciprocal of \frac{3392}{81}.

$$x=\frac{-45\sqrt{97}-75}{106}$$

The equation is now solved.

$$x=\frac{45\sqrt{97}-75}{106} x=\frac{-45\sqrt{97}-75}{106}$$

Steps for Completing the Square

Multiply both sides of the equation by 400, the least common multiple of 25,16.

$$16x^{2}+25\left(\frac{4}{9}x+\frac{4}{3}\right)^{2}=400$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\frac{4}{9}x+\frac{4}{3}\right)^{2}.

$$16x^{2}+25\left(\frac{16}{81}x^{2}+\frac{32}{27}x+\frac{16}{9}\right)=400$$

Use the distributive property to multiply 25 by \frac{16}{81}x^{2}+\frac{32}{27}x+\frac{16}{9}.

$$16x^{2}+\frac{400}{81}x^{2}+\frac{800}{27}x+\frac{400}{9}=400$$

Combine 16x^{2} and \frac{400}{81}x^{2} to get \frac{1696}{81}x^{2}.

$$\frac{1696}{81}x^{2}+\frac{800}{27}x+\frac{400}{9}=400$$

Subtract \frac{400}{9} from both sides.

$$\frac{1696}{81}x^{2}+\frac{800}{27}x=400-\frac{400}{9}$$

Subtract \frac{400}{9} from 400 to get \frac{3200}{9}.

$$\frac{1696}{81}x^{2}+\frac{800}{27}x=\frac{3200}{9}$$

Divide both sides of the equation by \frac{1696}{81}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{1696}{81}x^{2}+\frac{800}{27}x}{\frac{1696}{81}}=\frac{\frac{3200}{9}}{\frac{1696}{81}}$$

Dividing by \frac{1696}{81} undoes the multiplication by \frac{1696}{81}.

$$x^{2}+\frac{\frac{800}{27}}{\frac{1696}{81}}x=\frac{\frac{3200}{9}}{\frac{1696}{81}}$$

Divide \frac{800}{27} by \frac{1696}{81} by multiplying \frac{800}{27} by the reciprocal of \frac{1696}{81}.

$$x^{2}+\frac{75}{53}x=\frac{\frac{3200}{9}}{\frac{1696}{81}}$$

Divide \frac{3200}{9} by \frac{1696}{81} by multiplying \frac{3200}{9} by the reciprocal of \frac{1696}{81}.

$$x^{2}+\frac{75}{53}x=\frac{900}{53}$$

Divide \frac{75}{53}, the coefficient of the x term, by 2 to get \frac{75}{106}. Then add the square of \frac{75}{106} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{75}{53}x+\left(\frac{75}{106}\right)^{2}=\frac{900}{53}+\left(\frac{75}{106}\right)^{2}$$

Square \frac{75}{106} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{75}{53}x+\frac{5625}{11236}=\frac{900}{53}+\frac{5625}{11236}$$

Add \frac{900}{53} to \frac{5625}{11236} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{75}{53}x+\frac{5625}{11236}=\frac{196425}{11236}$$

Factor x^{2}+\frac{75}{53}x+\frac{5625}{11236}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{75}{106}\right)^{2}=\frac{196425}{11236}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{75}{106}\right)^{2}}=\sqrt{\frac{196425}{11236}}$$

Simplify.

$$x+\frac{75}{106}=\frac{45\sqrt{97}}{106} x+\frac{75}{106}=-\frac{45\sqrt{97}}{106}$$

Subtract \frac{75}{106} from both sides of the equation.

$$x=\frac{45\sqrt{97}-75}{106} x=\frac{-45\sqrt{97}-75}{106}$$