Solve {left(25/4+3/4xright)}^2+x^2=100

$${ \left( \frac{ 25 }{ 4 } + \frac{ 3 }{ 4 } x \right) }^{ 2 } + { x }^{ 2 } =100$$

Solve for x

$$x=4\sqrt{3}-3\approx 3.92820323$$

$$x=-4\sqrt{3}-3\approx -9.92820323$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\frac{25}{4}+\frac{3}{4}x\right)^{2}.

$$\frac{625}{16}+\frac{75}{8}x+\frac{9}{16}x^{2}+x^{2}=100$$

Combine \frac{9}{16}x^{2} and x^{2} to get \frac{25}{16}x^{2}.

$$\frac{625}{16}+\frac{75}{8}x+\frac{25}{16}x^{2}=100$$

Subtract 100 from both sides.

$$\frac{625}{16}+\frac{75}{8}x+\frac{25}{16}x^{2}-100=0$$

Subtract 100 from \frac{625}{16} to get -\frac{975}{16}.

$$-\frac{975}{16}+\frac{75}{8}x+\frac{25}{16}x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$\frac{25}{16}x^{2}+\frac{75}{8}x-\frac{975}{16}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{25}{16} for a, \frac{75}{8} for b, and -\frac{975}{16} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{75}{8}±\sqrt{\left(\frac{75}{8}\right)^{2}-4\times \frac{25}{16}\left(-\frac{975}{16}\right)}}{2\times \frac{25}{16}}$$

Square \frac{75}{8} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{75}{8}±\sqrt{\frac{5625}{64}-4\times \frac{25}{16}\left(-\frac{975}{16}\right)}}{2\times \frac{25}{16}}$$

Multiply -4 times \frac{25}{16}.

$$x=\frac{-\frac{75}{8}±\sqrt{\frac{5625}{64}-\frac{25}{4}\left(-\frac{975}{16}\right)}}{2\times \frac{25}{16}}$$

Multiply -\frac{25}{4} times -\frac{975}{16} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{75}{8}±\sqrt{\frac{5625+24375}{64}}}{2\times \frac{25}{16}}$$

Add \frac{5625}{64} to \frac{24375}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{75}{8}±\sqrt{\frac{1875}{4}}}{2\times \frac{25}{16}}$$

Take the square root of \frac{1875}{4}.

$$x=\frac{-\frac{75}{8}±\frac{25\sqrt{3}}{2}}{2\times \frac{25}{16}}$$

Multiply 2 times \frac{25}{16}.

$$x=\frac{-\frac{75}{8}±\frac{25\sqrt{3}}{2}}{\frac{25}{8}}$$

Now solve the equation x=\frac{-\frac{75}{8}±\frac{25\sqrt{3}}{2}}{\frac{25}{8}} when ± is plus. Add -\frac{75}{8} to \frac{25\sqrt{3}}{2}.

$$x=\frac{\frac{25\sqrt{3}}{2}-\frac{75}{8}}{\frac{25}{8}}$$

Divide -\frac{75}{8}+\frac{25\sqrt{3}}{2} by \frac{25}{8} by multiplying -\frac{75}{8}+\frac{25\sqrt{3}}{2} by the reciprocal of \frac{25}{8}.

$$x=4\sqrt{3}-3$$

Now solve the equation x=\frac{-\frac{75}{8}±\frac{25\sqrt{3}}{2}}{\frac{25}{8}} when ± is minus. Subtract \frac{25\sqrt{3}}{2} from -\frac{75}{8}.

$$x=\frac{-\frac{25\sqrt{3}}{2}-\frac{75}{8}}{\frac{25}{8}}$$

Divide -\frac{75}{8}-\frac{25\sqrt{3}}{2} by \frac{25}{8} by multiplying -\frac{75}{8}-\frac{25\sqrt{3}}{2} by the reciprocal of \frac{25}{8}.

$$x=-4\sqrt{3}-3$$

The equation is now solved.

$$x=4\sqrt{3}-3 x=-4\sqrt{3}-3$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\frac{25}{4}+\frac{3}{4}x\right)^{2}.

$$\frac{625}{16}+\frac{75}{8}x+\frac{9}{16}x^{2}+x^{2}=100$$

Combine \frac{9}{16}x^{2} and x^{2} to get \frac{25}{16}x^{2}.

$$\frac{625}{16}+\frac{75}{8}x+\frac{25}{16}x^{2}=100$$

Subtract \frac{625}{16} from both sides.

$$\frac{75}{8}x+\frac{25}{16}x^{2}=100-\frac{625}{16}$$

Subtract \frac{625}{16} from 100 to get \frac{975}{16}.

$$\frac{75}{8}x+\frac{25}{16}x^{2}=\frac{975}{16}$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$\frac{25}{16}x^{2}+\frac{75}{8}x=\frac{975}{16}$$

Divide both sides of the equation by \frac{25}{16}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{25}{16}x^{2}+\frac{75}{8}x}{\frac{25}{16}}=\frac{\frac{975}{16}}{\frac{25}{16}}$$

Dividing by \frac{25}{16} undoes the multiplication by \frac{25}{16}.

$$x^{2}+\frac{\frac{75}{8}}{\frac{25}{16}}x=\frac{\frac{975}{16}}{\frac{25}{16}}$$

Divide \frac{75}{8} by \frac{25}{16} by multiplying \frac{75}{8} by the reciprocal of \frac{25}{16}.

$$x^{2}+6x=\frac{\frac{975}{16}}{\frac{25}{16}}$$

Divide \frac{975}{16} by \frac{25}{16} by multiplying \frac{975}{16} by the reciprocal of \frac{25}{16}.

$$x^{2}+6x=39$$

Divide 6, the coefficient of the x term, by 2 to get 3. Then add the square of 3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+6x+3^{2}=39+3^{2}$$

Square 3.

$$x^{2}+6x+9=39+9$$

Add 39 to 9.

$$x^{2}+6x+9=48$$

Factor x^{2}+6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+3\right)^{2}=48$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{48}$$

Simplify.

$$x+3=4\sqrt{3} x+3=-4\sqrt{3}$$

Subtract 3 from both sides of the equation.

$$x=4\sqrt{3}-3 x=-4\sqrt{3}-3$$