Solve {left(3/4a+3right)}^2+a^2=16

$${ \left( \frac{ 3 }{ 4 } a+3 \right) }^{ 2 } + { a }^{ 2 } = 16$$

Solve for a

$$a=-4$$

$$a = \frac{28}{25} = 1\frac{3}{25} = 1.12$$

Steps Using the Quadratic Formula

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

$$\frac{9}{16}a^{2}+\frac{9}{2}a+9+a^{2}=16$$

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

$$\frac{25}{16}a^{2}+\frac{9}{2}a+9=16$$

Subtract 16 from both sides.

$$\frac{25}{16}a^{2}+\frac{9}{2}a+9-16=0$$

Subtract 16 from 9 to get -7.

$$\frac{25}{16}a^{2}+\frac{9}{2}a-7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{25}{16} for a, \frac{9}{2} for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-\frac{9}{2}±\sqrt{\left(\frac{9}{2}\right)^{2}-4\times \frac{25}{16}\left(-7\right)}}{2\times \frac{25}{16}}$$

Square \frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$a=\frac{-\frac{9}{2}±\sqrt{\frac{81}{4}-4\times \frac{25}{16}\left(-7\right)}}{2\times \frac{25}{16}}$$

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

$$a=\frac{-\frac{9}{2}±\sqrt{\frac{81}{4}-\frac{25}{4}\left(-7\right)}}{2\times \frac{25}{16}}$$

Multiply -\frac{25}{4} times -7.

$$a=\frac{-\frac{9}{2}±\sqrt{\frac{81+175}{4}}}{2\times \frac{25}{16}}$$

Add \frac{81}{4} to \frac{175}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a=\frac{-\frac{9}{2}±\sqrt{64}}{2\times \frac{25}{16}}$$

Take the square root of 64.

$$a=\frac{-\frac{9}{2}±8}{2\times \frac{25}{16}}$$

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

$$a=\frac{-\frac{9}{2}±8}{\frac{25}{8}}$$

Now solve the equation a=\frac{-\frac{9}{2}±8}{\frac{25}{8}} when ± is plus. Add -\frac{9}{2} to 8.

$$a=\frac{\frac{7}{2}}{\frac{25}{8}}$$

Divide \frac{7}{2} by \frac{25}{8} by multiplying \frac{7}{2} by the reciprocal of \frac{25}{8}.

$$a=\frac{28}{25}$$

Now solve the equation a=\frac{-\frac{9}{2}±8}{\frac{25}{8}} when ± is minus. Subtract 8 from -\frac{9}{2}.

$$a=-\frac{\frac{25}{2}}{\frac{25}{8}}$$

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

$$a=-4$$

The equation is now solved.

$$a=\frac{28}{25} a=-4$$

Steps for Completing the Square

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

$$\frac{9}{16}a^{2}+\frac{9}{2}a+9+a^{2}=16$$

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

$$\frac{25}{16}a^{2}+\frac{9}{2}a+9=16$$

Subtract 9 from both sides.

$$\frac{25}{16}a^{2}+\frac{9}{2}a=16-9$$

Subtract 9 from 16 to get 7.

$$\frac{25}{16}a^{2}+\frac{9}{2}a=7$$

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

$$\frac{\frac{25}{16}a^{2}+\frac{9}{2}a}{\frac{25}{16}}=\frac{7}{\frac{25}{16}}$$

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

$$a^{2}+\frac{\frac{9}{2}}{\frac{25}{16}}a=\frac{7}{\frac{25}{16}}$$

Divide \frac{9}{2} by \frac{25}{16} by multiplying \frac{9}{2} by the reciprocal of \frac{25}{16}.

$$a^{2}+\frac{72}{25}a=\frac{7}{\frac{25}{16}}$$

Divide 7 by \frac{25}{16} by multiplying 7 by the reciprocal of \frac{25}{16}.

$$a^{2}+\frac{72}{25}a=\frac{112}{25}$$

Divide \frac{72}{25}, the coefficient of the x term, by 2 to get \frac{36}{25}. Then add the square of \frac{36}{25} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}+\frac{72}{25}a+\left(\frac{36}{25}\right)^{2}=\frac{112}{25}+\left(\frac{36}{25}\right)^{2}$$

Square \frac{36}{25} by squaring both the numerator and the denominator of the fraction.

$$a^{2}+\frac{72}{25}a+\frac{1296}{625}=\frac{112}{25}+\frac{1296}{625}$$

Add \frac{112}{25} to \frac{1296}{625} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a^{2}+\frac{72}{25}a+\frac{1296}{625}=\frac{4096}{625}$$

Factor a^{2}+\frac{72}{25}a+\frac{1296}{625}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a+\frac{36}{25}\right)^{2}=\frac{4096}{625}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a+\frac{36}{25}\right)^{2}}=\sqrt{\frac{4096}{625}}$$

Simplify.

$$a+\frac{36}{25}=\frac{64}{25} a+\frac{36}{25}=-\frac{64}{25}$$

Subtract \frac{36}{25} from both sides of the equation.

$$a=\frac{28}{25} a=-4$$