# How do you solve 8y^2 - 22y + 15 = 0?

Jun 12, 2015

Use a modified AC Method to find: $8 {y}^{2} - 22 y + 15 = \left(2 y - 3\right) \left(4 y - 5\right)$, hence $y = \frac{3}{2}$ or $y = \frac{5}{4}$.

#### Explanation:

$f \left(y\right) = 8 {y}^{2} - 22 y + 15$ is of the form $a {y}^{2} + b y + c$ with $a = 8$, $b = - 22$ and $c = 15$.

Let $A = 8$, $B = 22$ and $C = 15$.

Look for a factorization of $A C = 8 \cdot 15 = 120$ into a pair of factors whose sum is $B = 22$.

Notice that $A + C = 23$, which is close to $22$, so the pair of factors we are looking for is close to $8 , 15$. Notice that $B 1 = 12$ and $B 2 = 10$ works.

Then for each pair $\left(A , B 1\right)$ and $\left(A , B 2\right)$, divide by the HCF (highest common factor) to get a pair of coefficients of a factor of $f \left(y\right)$ (choosing suitable signs) as follows:

$\left(A , B 1\right) = \left(8 , 12\right) \to \left(2 , 3\right) \to \left(2 y - 3\right)$
$\left(A , B 2\right) = \left(8 , 10\right) \to \left(4 , 5\right) \to \left(4 y - 5\right)$

Hence $8 {y}^{2} - 22 y + 15 = \left(2 y - 3\right) \left(4 y - 5\right)$

If $f \left(y\right) = 0$ then $\left(2 y - 3\right) = 0$ giving $y = \frac{3}{2}$

or $\left(4 y - 5\right) = 0$ giving $y = \frac{5}{4}$.