Example 3
Determine the roots.
- \(8x^2+10x=3\)
- \(x^2+18=11x\)
Include a check for part b).
Before we start, let's review our methods for solving quadratic equations not written in vertex form.
Solving Quadratic Equations (not written in vertex form)
- Write the equation in the form \(ax^2+bx+c=0\).
In other words, rearrange the equation so that it equals \(0\).
- Choose: Factor or Quadratic Formula
- Factor: more efficient (except perhaps if decomposition is needed), but factoring will not find roots that are irrational
- Quadratic Formula, \(x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\) always works if the roots exist
Note that isolating for \(x\) is not an option when there are both \(x^2\)- and \(x\)-terms. This is because the two terms involving \(x\) cannot be gathered together. Consequently, it's not possible to isolate for \(x\) when there are \(x^2\)- and \(x\)-terms.
Solution — Part A
Recall part a): Determine the roots of \(8x^2+10x=3\).
Step 1: Write the equation in the form \(ax^2+bx+c=0\).
\( 8 x^{2}+10 x =3 \)
In this case, we need to get rid of the \(3\) from the right side.
We can do that by subtracting \(3\) from both sides.
\(\begin{align*} 8 x^{2}+10 x-3 &=3-3 \\ 8 x^{2}+10 x-3 &=0 \end{align*}\)
Now we decide to either factor or use the quadratic formula.
Since this equation would require decomposition, we will use the quadratic formula instead.
Step 2: Use the quadratic formula with \(a=8\), \(b=10\), and \(c=-3\):
Notice that these are taken from the rearranged equation. We substitute these values into the formula.
\(\begin{align*} x &=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\ &=\frac{-10 \pm \sqrt{10^{2}-4(8)(-3)}}{2(8)} \end{align*}\)
When evaluating, first evaluate the radicant
\(\begin{align*} x &=\frac{-10 \pm \sqrt{196}}{16} \\ &=\frac{-10 \pm 14}{-16} \end{align*}\)
In future examples, we will see what to do when the square root is a decimal.
Finally, determine the two solutions by evaluating the addition and subtraction separately.
\(\begin{align*} x &=\dfrac{-10+14}{16} \\ x &=\dfrac{4}{16} \end{align*}\)
\(\begin{align*} x &=\frac{-10-14}{16} \\ x &=\frac{-24}{16} \end{align*}\)
After evaluating, write answers as fractions in lowest terms.
Recall
Check your answers by substituting them into the original equation!
Solution — Part B
Recall part b): Determine the roots of \(x^2+18=11x\), include a check after your solution.
The \(x^2\)-term reminds us that this is a quadratic equation.
Step 1: Write the equation in the form \(ax^2+bx+c=0\)
We do this by subtracting \(11x\) from both sides,
\(x^2-11x+18=0\)
Factor or Formula? Since the coefficient of \(x^2\) is \(1\), factoring will be quicker as it can be done in one step.
Step 2: Factor
We require a pair of integers that
- add to \(-11\),
- and multiply to \(18\).
The integers are \(-2\) and \(-9\).
So the equation becomes
\(\begin{array}{l}{x^{2}-11 x+18=0} \\ {(x-2)(x-9)=0}\end{array}\)
The roots are the values of \(x\) that make either factor \(0\).
Therefore, \(x=2\) or \(x=9\).
Recall
Check your answers by substituting them into the original equation!
There are two roots, so we need to check them both.
Check if \(x=2\) and \(x=9\) are solutions to the equation \(x^2+18=11x\).
Going back to the original equation, separate it into its left side, \(x^2 + 18\), and its right side, \(11x\).
We may not write this as an equation anymore, because we should not claim that the two sides are equal, since that is what we are checking.
Check \(x=2\):
\(\begin{align*} L S &=x^{2}+18 \\ &=(2)^{2}+18 \\ &=22 \end{align*}\)
\(\begin{align*} R S &=11 x \\ &=11(2) \\ &=22 \end{align*}\)
Therefore, \(LS=RS\).
Check \(x=9\):
\(\begin{align*} L S &=x^{2}+18 \\ &=9^{2}+18 \\ &=99 \end{align*}\)
\(\begin{align*} R S &=11 x \\ &=11(9) \\ &=99 \end{align*}\)
Therefore, \(LS=RS\).