Solve ((a-1)/2)^2=a^2-7a-4

$$( \frac { ( a - 1 ) } { 2 } ) ^ { 2 } = a ^ { 2 } - 7 a - 4$$

Solve for a

$$a = \frac{2 \sqrt{55} + 13}{3} \approx 9.277465658$$

$$a=\frac{13-2\sqrt{55}}{3}\approx -0.610798991$$

Steps Using the Quadratic Formula

To raise \frac{a-1}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(a-1\right)^{2}}{2^{2}}=a^{2}-7a-4$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(a-1\right)^{2}.

$$\frac{a^{2}-2a+1}{2^{2}}=a^{2}-7a-4$$

Calculate 2 to the power of 2 and get 4.

$$\frac{a^{2}-2a+1}{4}=a^{2}-7a-4$$

Divide each term of a^{2}-2a+1 by 4 to get \frac{1}{4}a^{2}-\frac{1}{2}a+\frac{1}{4}.

$$\frac{1}{4}a^{2}-\frac{1}{2}a+\frac{1}{4}=a^{2}-7a-4$$

Subtract a^{2} from both sides.

$$\frac{1}{4}a^{2}-\frac{1}{2}a+\frac{1}{4}-a^{2}=-7a-4$$

Combine \frac{1}{4}a^{2} and -a^{2} to get -\frac{3}{4}a^{2}.

$$-\frac{3}{4}a^{2}-\frac{1}{2}a+\frac{1}{4}=-7a-4$$

Add 7a to both sides.

$$-\frac{3}{4}a^{2}-\frac{1}{2}a+\frac{1}{4}+7a=-4$$

Combine -\frac{1}{2}a and 7a to get \frac{13}{2}a.

$$-\frac{3}{4}a^{2}+\frac{13}{2}a+\frac{1}{4}=-4$$

Add 4 to both sides.

$$-\frac{3}{4}a^{2}+\frac{13}{2}a+\frac{1}{4}+4=0$$

Add \frac{1}{4} and 4 to get \frac{17}{4}.

$$-\frac{3}{4}a^{2}+\frac{13}{2}a+\frac{17}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{3}{4} for a, \frac{13}{2} for b, and \frac{17}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-\frac{13}{2}±\sqrt{\left(\frac{13}{2}\right)^{2}-4\left(-\frac{3}{4}\right)\times \frac{17}{4}}}{2\left(-\frac{3}{4}\right)}$$

Square \frac{13}{2} by squaring both the numerator and the denominator of the fraction.

$$a=\frac{-\frac{13}{2}±\sqrt{\frac{169}{4}-4\left(-\frac{3}{4}\right)\times \frac{17}{4}}}{2\left(-\frac{3}{4}\right)}$$

Multiply -4 times -\frac{3}{4}.

$$a=\frac{-\frac{13}{2}±\sqrt{\frac{169}{4}+3\times \frac{17}{4}}}{2\left(-\frac{3}{4}\right)}$$

Multiply 3 times \frac{17}{4}.

$$a=\frac{-\frac{13}{2}±\sqrt{\frac{169+51}{4}}}{2\left(-\frac{3}{4}\right)}$$

Add \frac{169}{4} to \frac{51}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a=\frac{-\frac{13}{2}±\sqrt{55}}{2\left(-\frac{3}{4}\right)}$$

Multiply 2 times -\frac{3}{4}.

$$a=\frac{-\frac{13}{2}±\sqrt{55}}{-\frac{3}{2}}$$

Now solve the equation a=\frac{-\frac{13}{2}±\sqrt{55}}{-\frac{3}{2}} when ± is plus. Add -\frac{13}{2} to \sqrt{55}.

$$a=\frac{\sqrt{55}-\frac{13}{2}}{-\frac{3}{2}}$$

Divide -\frac{13}{2}+\sqrt{55} by -\frac{3}{2} by multiplying -\frac{13}{2}+\sqrt{55} by the reciprocal of -\frac{3}{2}.

$$a=\frac{13-2\sqrt{55}}{3}$$

Now solve the equation a=\frac{-\frac{13}{2}±\sqrt{55}}{-\frac{3}{2}} when ± is minus. Subtract \sqrt{55} from -\frac{13}{2}.

$$a=\frac{-\sqrt{55}-\frac{13}{2}}{-\frac{3}{2}}$$

Divide -\frac{13}{2}-\sqrt{55} by -\frac{3}{2} by multiplying -\frac{13}{2}-\sqrt{55} by the reciprocal of -\frac{3}{2}.

$$a=\frac{2\sqrt{55}+13}{3}$$

The equation is now solved.

$$a=\frac{13-2\sqrt{55}}{3} a=\frac{2\sqrt{55}+13}{3}$$

Steps for Completing the Square

To raise \frac{a-1}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(a-1\right)^{2}}{2^{2}}=a^{2}-7a-4$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(a-1\right)^{2}.

$$\frac{a^{2}-2a+1}{2^{2}}=a^{2}-7a-4$$

Calculate 2 to the power of 2 and get 4.

$$\frac{a^{2}-2a+1}{4}=a^{2}-7a-4$$

Divide each term of a^{2}-2a+1 by 4 to get \frac{1}{4}a^{2}-\frac{1}{2}a+\frac{1}{4}.

$$\frac{1}{4}a^{2}-\frac{1}{2}a+\frac{1}{4}=a^{2}-7a-4$$

Subtract a^{2} from both sides.

$$\frac{1}{4}a^{2}-\frac{1}{2}a+\frac{1}{4}-a^{2}=-7a-4$$

Combine \frac{1}{4}a^{2} and -a^{2} to get -\frac{3}{4}a^{2}.

$$-\frac{3}{4}a^{2}-\frac{1}{2}a+\frac{1}{4}=-7a-4$$

Add 7a to both sides.

$$-\frac{3}{4}a^{2}-\frac{1}{2}a+\frac{1}{4}+7a=-4$$

Combine -\frac{1}{2}a and 7a to get \frac{13}{2}a.

$$-\frac{3}{4}a^{2}+\frac{13}{2}a+\frac{1}{4}=-4$$

Subtract \frac{1}{4} from both sides.

$$-\frac{3}{4}a^{2}+\frac{13}{2}a=-4-\frac{1}{4}$$

Subtract \frac{1}{4} from -4 to get -\frac{17}{4}.

$$-\frac{3}{4}a^{2}+\frac{13}{2}a=-\frac{17}{4}$$

Divide both sides of the equation by -\frac{3}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{3}{4}a^{2}+\frac{13}{2}a}{-\frac{3}{4}}=-\frac{\frac{17}{4}}{-\frac{3}{4}}$$

Dividing by -\frac{3}{4} undoes the multiplication by -\frac{3}{4}.

$$a^{2}+\frac{\frac{13}{2}}{-\frac{3}{4}}a=-\frac{\frac{17}{4}}{-\frac{3}{4}}$$

Divide \frac{13}{2} by -\frac{3}{4} by multiplying \frac{13}{2} by the reciprocal of -\frac{3}{4}.

$$a^{2}-\frac{26}{3}a=-\frac{\frac{17}{4}}{-\frac{3}{4}}$$

Divide -\frac{17}{4} by -\frac{3}{4} by multiplying -\frac{17}{4} by the reciprocal of -\frac{3}{4}.

$$a^{2}-\frac{26}{3}a=\frac{17}{3}$$

Divide -\frac{26}{3}, the coefficient of the x term, by 2 to get -\frac{13}{3}. Then add the square of -\frac{13}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}-\frac{26}{3}a+\left(-\frac{13}{3}\right)^{2}=\frac{17}{3}+\left(-\frac{13}{3}\right)^{2}$$

Square -\frac{13}{3} by squaring both the numerator and the denominator of the fraction.

$$a^{2}-\frac{26}{3}a+\frac{169}{9}=\frac{17}{3}+\frac{169}{9}$$

Add \frac{17}{3} to \frac{169}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a^{2}-\frac{26}{3}a+\frac{169}{9}=\frac{220}{9}$$

Factor a^{2}-\frac{26}{3}a+\frac{169}{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}.

$$\left(a-\frac{13}{3}\right)^{2}=\frac{220}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-\frac{13}{3}\right)^{2}}=\sqrt{\frac{220}{9}}$$

Simplify.

$$a-\frac{13}{3}=\frac{2\sqrt{55}}{3} a-\frac{13}{3}=-\frac{2\sqrt{55}}{3}$$

Add \frac{13}{3} to both sides of the equation.

$$a=\frac{2\sqrt{55}+13}{3} a=\frac{13-2\sqrt{55}}{3}$$