# Question #9a872

Jan 20, 2018

The graphs of the two equations intersect at the points $\left(6 , 23\right)$ and $\left(3 , 14\right)$

#### Explanation:

The formatting seems to have taken a hit there, so we'll assume that the equations are

$y = {x}^{2} - 6 x + 23$ and
$y - 5 = 3 x$.

The second equation can be written as $y = 3 x + 5$. Both equations need to be true for specific values of $x$ and $y$, so we can substitute the value of $y$ in terms of $x$ in the first equation to solve for $x$:

$3 x + 5 = {x}^{2} - 6 x + 23 \implies {x}^{2} - 9 x + 18 = 0$

This is just a quadratic equation, so let's find the discriminant:

$D = {\left(- 9\right)}^{2} - 4 \cdot 1 \cdot 18 = 81 - 72 = 9 > 0$

Since the discriminant is greater than $0$, we know the equation has two distinct roots.

${x}_{1} = \frac{9 + 3}{2} = 6$

${x}_{2} = \frac{9 - 3}{2} = 3$

Then we just need to use those values of $x$ to determine $y$ using $y = 3 x + 5$. If $x = 6$, $y = 23$ and if $x = 3$, $y = 14$

Therefore, the graphs of the two equations intersect at the points $\left(6 , 23\right)$ and $\left(3 , 14\right)$

NOTE: For any equation of the form $a {x}^{2} + b x + c = 0$, the discriminant is calculated by $D = {b}^{2} - 4 a c$ and the roots are calculated by $\frac{- b \pm \sqrt{D}}{2 a}$ if $D \ge 0$, and if $D < 0$ there are no real roots.