Question

How do you use the quadratic formula to solve for $x$-intercepts ${x^2} - 8x + 12 = 0?$

Answer 1

Hint: In this question using the quadratic formula. First we take the given equation. We identify the value of $a, b$ and $c$ in the quadratic equation. After that Substitute the values $a, b$ and $c$ into the quadratic formula and solve for $x$. Now simplify the equation, hence we get the equation.
Use the quadratic formula to find the solutions,
\[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Pull terms out from under the radical, assuming positive real numbers. Now we find the $x$ value. First we separate the positive and sign values.
Finally we get $x$ values.

Complete step by step answer:
The given quadratic equation is ${x^2} - 8x + 12 = 0$
The quadratic formula states:
For $a{x^2} + bx + c = 0$, the value of $x$ which are the solutions to the equation are given by:
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Where, $a = 1,\,b = - 8$ and $c = 12$
Now, we substitute these values in the quadratic formula
$x = \dfrac{{ - ( - 8) \pm \sqrt {{{( - 8)}^2} - 4(1)(12)} }}{{2(1)}}$
We simplify the equation, hence we get
$x = \dfrac{{(8) \pm \sqrt {{{( - 8)}^2} - 4(1)(12)} }}{{2(1)}}$
Raise$ - 8$ to the power of $2$
$x = \dfrac{{(8) \pm \sqrt {(64) - 4(1)(12)} }}{{2(1)}}$
Multiply $1$by $12$
$x = \dfrac{{(8) \pm \sqrt {(64) - 4 \times 12} }}{{2(1)}}$
Multiply $4$ by $2$
\[x = \dfrac{{(8) \pm \sqrt {64 - 48} }}{{2(1)}}\]
Subtract $48$ from $64$
\[x = \dfrac{{(8) \pm \sqrt {16} }}{{2(1)}}\]
Rewrite $16$ as ${4^2}$
\[x = \dfrac{{(8) \pm \sqrt {{4^2}} }}{{2(1)}}\]
Pull terms out from under the radical, assuming positive real numbers
\[x = \dfrac{{8 \pm 4}}{{2 \times 1}}\]
Multiply $2$ by $1$
\[x = \dfrac{{8 \pm 4}}{2}\]
Now we find the $x$ value. First we separate the positive and negative sign values,
Let,
\[x = \dfrac{{8 + 4}}{2}\]
Add the numerator, hence we get
\[x = \dfrac{{12}}{2}\]
Divide $12$ by $2$
$x = 6$
Let,
\[x = \dfrac{{8 - 4}}{2}\]
Subtract the numerator
\[x = \dfrac{4}{2}\]
Divide $4$ by $2$
$x = 2$
The given equation solution is $6$ and $2$.
The values of $x = 6,2$

Note: The formula for finding the roots of the quadratic equation $a{x^2} + bx + c = 0$ is \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
The formula for finding roots of a quadratic equation was known to ancient Babylonians, though not in a form as we derived.
They found the roots by creating the steps as a verse, which is a common practice at their times.
Babylonians used quadratic equations for deciding to choose the dimensions of their land for agriculture.