Solve y^2=9y-8

$$y ^ { 2 } = 9 y - 8$$

Solve for y

$$y=1$$

$$y=8$$

Steps Using Factoring

Subtract 9y from both sides.

$$y^{2}-9y=-8$$

Add 8 to both sides.

$$y^{2}-9y+8=0$$

To solve the equation, factor y^{2}-9y+8 using formula y^{2}+\left(a+b\right)y+ab=\left(y+a\right)\left(y+b\right). To find a and b, set up a system to be solved.

$$a+b=-9 ab=8$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 8.

$$-1,-8 -2,-4$$

Calculate the sum for each pair.

$$-1-8=-9 -2-4=-6$$

The solution is the pair that gives sum -9.

$$a=-8 b=-1$$

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

$$\left(y-8\right)\left(y-1\right)$$

To find equation solutions, solve y-8=0 and y-1=0.

$$y=8 y=1$$

Steps Using Factoring By Grouping

Subtract 9y from both sides.

$$y^{2}-9y=-8$$

Add 8 to both sides.

$$y^{2}-9y+8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by+8. To find a and b, set up a system to be solved.

$$a+b=-9 ab=1\times 8=8$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 8.

$$-1,-8 -2,-4$$

Calculate the sum for each pair.

$$-1-8=-9 -2-4=-6$$

The solution is the pair that gives sum -9.

$$a=-8 b=-1$$

Rewrite y^{2}-9y+8 as \left(y^{2}-8y\right)+\left(-y+8\right).

$$\left(y^{2}-8y\right)+\left(-y+8\right)$$

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

$$y\left(y-8\right)-\left(y-8\right)$$

Factor out common term y-8 by using distributive property.

$$\left(y-8\right)\left(y-1\right)$$

To find equation solutions, solve y-8=0 and y-1=0.

$$y=8 y=1$$

Steps Using the Quadratic Formula

Subtract 9y from both sides.

$$y^{2}-9y=-8$$

Add 8 to both sides.

$$y^{2}-9y+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -9 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 8}}{2}$$

Square -9.

$$y=\frac{-\left(-9\right)±\sqrt{81-4\times 8}}{2}$$

Multiply -4 times 8.

$$y=\frac{-\left(-9\right)±\sqrt{81-32}}{2}$$

Add 81 to -32.

$$y=\frac{-\left(-9\right)±\sqrt{49}}{2}$$

Take the square root of 49.

$$y=\frac{-\left(-9\right)±7}{2}$$

The opposite of -9 is 9.

$$y=\frac{9±7}{2}$$

Now solve the equation y=\frac{9±7}{2} when ± is plus. Add 9 to 7.

$$y=\frac{16}{2}$$

Divide 16 by 2.

$$y=8$$

Now solve the equation y=\frac{9±7}{2} when ± is minus. Subtract 7 from 9.

$$y=\frac{2}{2}$$

Divide 2 by 2.

$$y=1$$

The equation is now solved.

$$y=8 y=1$$

Steps for Completing the Square

Subtract 9y from both sides.

$$y^{2}-9y=-8$$

Divide -9, the coefficient of the x term, by 2 to get -\frac{9}{2}. Then add the square of -\frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-9y+\left(-\frac{9}{2}\right)^{2}=-8+\left(-\frac{9}{2}\right)^{2}$$

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

$$y^{2}-9y+\frac{81}{4}=-8+\frac{81}{4}$$

Add -8 to \frac{81}{4}.

$$y^{2}-9y+\frac{81}{4}=\frac{49}{4}$$

Factor y^{2}-9y+\frac{81}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-\frac{9}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{9}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$y-\frac{9}{2}=\frac{7}{2} y-\frac{9}{2}=-\frac{7}{2}$$

Add \frac{9}{2} to both sides of the equation.

$$y=8 y=1$$