Solve b^2=8b-7

$$b ^ { 2 } = 8 b - 7$$

Solve for b

$$b=1$$

$$b=7$$

Steps Using Factoring

Subtract 8b from both sides.

$$b^{2}-8b=-7$$

Add 7 to both sides.

$$b^{2}-8b+7=0$$

To solve the equation, factor b^{2}-8b+7 using formula b^{2}+\left(a+b\right)b+ab=\left(b+a\right)\left(b+b\right). To find a and b, set up a system to be solved.

$$a+b=-8 ab=7$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-7 b=-1$$

Rewrite factored expression \left(b+a\right)\left(b+b\right) using the obtained values.

$$\left(b-7\right)\left(b-1\right)$$

To find equation solutions, solve b-7=0 and b-1=0.

$$b=7 b=1$$

Steps Using Factoring By Grouping

Subtract 8b from both sides.

$$b^{2}-8b=-7$$

Add 7 to both sides.

$$b^{2}-8b+7=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as b^{2}+ab+bb+7. To find a and b, set up a system to be solved.

$$a+b=-8 ab=1\times 7=7$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-7 b=-1$$

Rewrite b^{2}-8b+7 as \left(b^{2}-7b\right)+\left(-b+7\right).

$$\left(b^{2}-7b\right)+\left(-b+7\right)$$

Factor out b in the first and -1 in the second group.

$$b\left(b-7\right)-\left(b-7\right)$$

Factor out common term b-7 by using distributive property.

$$\left(b-7\right)\left(b-1\right)$$

To find equation solutions, solve b-7=0 and b-1=0.

$$b=7 b=1$$

Steps Using the Quadratic Formula

Subtract 8b from both sides.

$$b^{2}-8b=-7$$

Add 7 to both sides.

$$b^{2}-8b+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -8 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$b=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 7}}{2}$$

Square -8.

$$b=\frac{-\left(-8\right)±\sqrt{64-4\times 7}}{2}$$

Multiply -4 times 7.

$$b=\frac{-\left(-8\right)±\sqrt{64-28}}{2}$$

Add 64 to -28.

$$b=\frac{-\left(-8\right)±\sqrt{36}}{2}$$

Take the square root of 36.

$$b=\frac{-\left(-8\right)±6}{2}$$

The opposite of -8 is 8.

$$b=\frac{8±6}{2}$$

Now solve the equation b=\frac{8±6}{2} when ± is plus. Add 8 to 6.

$$b=\frac{14}{2}$$

Divide 14 by 2.

$$b=7$$

Now solve the equation b=\frac{8±6}{2} when ± is minus. Subtract 6 from 8.

$$b=\frac{2}{2}$$

Divide 2 by 2.

$$b=1$$

The equation is now solved.

$$b=7 b=1$$

Steps for Completing the Square

Subtract 8b from both sides.

$$b^{2}-8b=-7$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$b^{2}-8b+\left(-4\right)^{2}=-7+\left(-4\right)^{2}$$

Square -4.

$$b^{2}-8b+16=-7+16$$

Add -7 to 16.

$$b^{2}-8b+16=9$$

Factor b^{2}-8b+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(b-4\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(b-4\right)^{2}}=\sqrt{9}$$

Simplify.

$$b-4=3 b-4=-3$$

Add 4 to both sides of the equation.

$$b=7 b=1$$