Common Calculation Errors When Using the Quadratic Formula

9.7.2 Common Calculation Errors When Using the Quadratic Formula

9.7.2 • Common Calculation Errors When Using the Quadratic Formula

Activity

Here are four equations, followed by attempts to solve them using the quadratic formula. Each attempt contains at least one error.

Solve the equations by using the quadratic formula. Then, find and describe the error(s) in the worked solutions of the same equations as the ones you solved.

Equation 1: $2 x^{2} + 3 = 8 x$

Here is the worked solution with errors:

Step 1 - $2 x^{2} - 8 x + 3 = 0$

Step 2 - $a = 2$ , $b = - 8$ , $c = 3$

Step 3 - $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

Step 4 - $x = \frac{- (- 8) \pm \sqrt{(- 8)^{2} - (4) (2) (3)}}{2 (2)}$

Step 5 - $x = \frac{8 \pm \sqrt{64 - 24}}{4}$

Step 6 - $x = \frac{8 \pm \sqrt{40}}{4}$

Step 7 - $x = 2 \pm \sqrt{10}$

Compare your answer:

The correct solutions are $x = 2 \pm \frac{\sqrt{40}}{4}$ or $x = 2 \pm \frac{1}{2} \sqrt{10}$ or equivalent.
The computation from Step 6 to Step 7 is not evaluated correctly, because the square root of the discriminant is incorrectly simplified. $\frac{\sqrt{40}}{4}$ is not $\sqrt{10}$ , it is $\frac{\sqrt{4} \cdot \sqrt{10}}{4} = \frac{2 \sqrt{10}}{4} = \frac{\sqrt{10}}{2}$ . Thus, the solution is wrong.

$\begin{matrix} 2 x^{2} - 8 x + 3 = 0 \\ a = 2, b = - 8, c = 3 \\ x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ x = \frac{- (- 8) \pm \sqrt{(- 8)^{2} - (4) (2) (3)}}{2 (2)} \\ x = \frac{8 \pm \sqrt{64 - 24}}{4} \\ x = \frac{8 \pm \sqrt{40}}{4} \\ x = 2 \pm \frac{\sqrt{40}}{4} \\ x = 2 \pm \frac{\sqrt{4} \cdot \sqrt{10}}{4} \\ x = 2 \pm \frac{2 \cdot \sqrt{10}}{4} \\ x = 2 \pm \frac{1}{2} \sqrt{10} \end{matrix}$

Equation 2: $x^{2} + 3 x = 10$

Here is the worked solution with errors:

Step 1 - $a = 1$ , $b = 3$ , $c = 10$

Step 2 - $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

Step 3 - $x = \frac{- (3) \pm \sqrt{3^{2} - (4) (1) (10)}}{2 (1)}$

Step 4 - $x = \frac{- 3 \pm \sqrt{9 - 40}}{2}$

Step 5 - $x = \frac{- 3 \pm \sqrt{- 31}}{2}$

No Solutions

Compare your answer:

The correct solutions are $x = 2$ and $x = - 5$ .
The equation was not rewritten in standard form, =0, before Step 1, where the $a$ , $b$ , and $c$ were identified. The $c$ should be -10, not 10.

$\begin{matrix} x^{2} + 3 x - 10 = 0 \\ a = 1, b = 3, c = - 10 \\ x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ x = \frac{- (3) \pm \sqrt{(3)^{2} - (4) (1) (- 10)}}{2 (1)} \\ x = \frac{- 3 \pm \sqrt{9 - (- 40)}}{2} \\ x = \frac{- 3 \pm \sqrt{9 + 40}}{2} \\ x = \frac{- 3 \pm \sqrt{49}}{2} \\ x = \frac{- 3 \pm 7}{2} \\ x = \frac{- 3 + 7}{2} = \frac{4}{2} = 2 \\ x = \frac{- 3 - 7}{2} = \frac{- 10}{2} = - 5 \end{matrix}$

Equation 3: $9 x^{2} - 2 x - 1 = 0$

Here is the worked solution with errors:

Step 1 - $a = 9$ , $b = - 2$ , $c = - 1$

Step 2 - $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

Step 3 - $x = \frac{2 \pm \sqrt{(- 2)^{2} - (4) (9) (- 1)}}{2}$

Step 4 - $x = \frac{2 \pm \sqrt{4 + 36}}{2}$

Step 5 - $x = \frac{2 \pm \sqrt{40}}{2}$

Compare your answer:

The correct solution is $x = \frac{1 \pm \sqrt{10}}{9}$ (or equivalent).
In step 3, the value of $a$ is not multiplied by 2 in the denominator of the formula.

$\begin{matrix} a = 9, b = - 2, c = - 1 \\ x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ x = \frac{- (- 2) \pm \sqrt{(- 2)^{2} - (4) (9) (- 1)}}{2 (9)} \\ x = \frac{2 \pm \sqrt{4 + 36}}{18} \\ x = \frac{2 \pm \sqrt{40}}{18} \\ x = \frac{2 \pm \sqrt{4} \cdot \sqrt{10}}{18} \\ x = \frac{2 \pm 2 \cdot \sqrt{10}}{2 (9)} \\ x = \frac{1 \pm \sqrt{10}}{9} \end{matrix}$

Equation 4: $x^{2} - 10 x + 23 = 0$

Here is the worked solution with errors:

Step 1 - $a = 1$ , $b = - 10$ , $c = 23$

Step 2 - $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

Step 3 - $x = \frac{- 10 \pm \sqrt{(- 10)^{2} - (4) (1) (23)}}{2}$

Step 4 - $x = \frac{- 10 \pm \sqrt{- 100 - 92}}{2}$

Step 5 - $x = \frac{- 10 \pm \sqrt{- 192}}{2}$

No Solutions

Compare your answer:

The correct solution is $x = 5 \pm \sqrt{2}$ (or equivalent).
In Step 3, when $b$ is -10, the value of $- b$ should be 10.
In Step 4, the value of $b^{2}$ should be positive even when $b$ is negative.

$\begin{matrix} a = 1, b = - 10, c = 23 \\ x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ x = \frac{- (- 10) \pm \sqrt{(- 10)^{2} - (4) (1) (23)}}{2} \\ x = \frac{10 \pm \sqrt{100 - 92}}{2} \\ x = \frac{10 \pm \sqrt{8}}{2} \\ x = 5 \pm \frac{\sqrt{8}}{2} \\ x = 5 \pm \frac{\sqrt{4} \cdot \sqrt{2}}{2} \\ x = 5 \pm \frac{2 \cdot \sqrt{2}}{2} \\ x = 5 \pm \sqrt{2} \end{matrix}$

Self Check

When solving $x^2−30x+26=0$ using the quadratic formula, Sloane's work showed the formula as this:

$x=\frac{-30 \pm \sqrt {900-104}}{2}$ .

Which of the following describes the error in her work?

The discriminant should be $-900 - 104$ , not $900 - 104$ ; $x=\frac {-30 \pm \sqrt{900-104}}{2}$ .
The first term in the numerator should be 30, not -30; $x=\frac {30 \pm \sqrt{900-104}}{2}$ .
The discriminant should be $900 + 104$ , not $900 - 104$ ; $x=\frac {-30 \pm \sqrt{900+104}}{2}$ .
The denominator should be 4, not 2; $x=\frac {-30 \pm \sqrt{900-104}}{4}$ .

Additional Resources

Common Errors When Using the Quadratic Formula

The quadratic formula has many parts in it. A small error in any one part can lead to incorrect solutions.

Suppose we are solving $2 x^{2} - 6 = 11 x$ . To use the formula, let's rewrite it in the form of $a x^{2} + b x + c = 0$ , which gives: $2 x^{2} - 11 x - 6 = 0$ .

Here are some common errors to avoid:

Using the wrong values for $a$ , $b$ , and $c$ in the formula.

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

$x = \frac{- 11 \pm \sqrt{(11)^{2} - 4 (2) (- 6)}}{2 (2)}$

This is incorrect! $b$ is -11, so $- b$ is -(-11), which is 11, not -11.

The correct formula is:
$x = \frac{11 \pm \sqrt{(11)^{2} - 4 (2) (- 6)}}{2 (2)}$

Forgetting to multiply 2 by $a$ for the denominator in the formula.

$x = \frac{11 \pm \sqrt{(11)^{2} - 4 (2) (- 6)}}{2}$

This is incorrect! The denominator is $2 a$ , which is $2 (2)$ or 4.

The correct formula is:
$x = \frac{11 \pm \sqrt{(11)^{2} - 4 (2) (- 6)}}{2 (2)}$

Forgetting that squaring a negative number produces a positive number.

$x = \frac{11 \pm \sqrt{- 121 - 4 (2) (- 6)}}{4}$

This is incorrect! $(- 11)^{2}$ is 121, not -121.

The correct formula is:
$x = \frac{11 \pm \sqrt{121 - 4 (2) (- 6)}}{4}$

Forgetting that a negative number times a positive number is a negative number.

$x = \frac{11 \pm \sqrt{121 - 48}}{4}$

This is incorrect! $4 (2) (- 6) = - 48$ and $121 - (- 48)$ is $121 + 48$ .

The correct formula is:
$x = \frac{11 \pm \sqrt{121 + 48}}{4}$

Making calculation errors or not following the properties of algebra.

$x = \frac{11 \pm \sqrt{169}}{4}$

$x = 11 \pm \sqrt{42.25}$

This is inBoth parts of the numerator, the 11 and the $\sqrt{169}$ , get divided by 4. Also, $\frac{\sqrt{169}}{4}$ is not $\sqrt{42.25}$ .

The correct formula is:
$x = \frac{11 \pm 13}{4}$

Let's finish by evaluating $\frac{11 \pm 13}{4}$ correctly:
$x = \frac{11 + 13}{4}$ or $x = \frac{11 - 13}{4}$
$x = \frac{24}{4}$ or $x = - \frac{2}{4}$
$x = 6$ or $x = - \frac{1}{2}$

To make sure our solutions are indeed correct, we can substitute the solutions back into the original equations and see whether each solution keeps the equation true.

Checking 6 as a solution:
$2 x^{2} - 6 = 11 x$
$2 (6)^{2} - 6 = 11 (6)$
$2 (36) - 6 = 66$
$72 - 6 = 66$ $66 = 66 ✓$

Checking $- \frac{1}{2}$ as a solution:
$2 x^{2} - 6 = 11 x$
$2 (- \frac{1}{2})^{2} - 6 = 11 (- \frac{1}{2})$
$2 (\frac{1}{4}) - 6 = - \frac{11}{2}$
$\frac{1}{2} - 6 = - 5 \frac{1}{2}$
$- 5 \frac{1}{2} = - 5 \frac{1}{2} ✓$

We can also graph the equation $y = 2 x^{2} - 11 x - 6$ and find its $x$ -intercepts to see whether our solutions to $2 x^{2} - 11 x - 6 = 0$ are accurate (or close to accurate).

Graph of a parabola on a coordinate plane. Two points (negative one-half, 0) and (6, 0) have been labeled on the parabola. The x-axis has a scale of 1 extending from negative 3 to 8. The y-axis extends from negative 24 to 8 with a scale of 4.

Try it

Common Errors When Using the Quadratic Formula

Zariah was solving the equation $3 x^{2} - 8 x + 4 = 0$ with the quadratic formula and used the steps below. She then had to stop because she realized there was a problem. Locate and describe her error.

Step 1 - $x = \frac{- (- 8) \pm \sqrt{- 8^{2} - 4 (3) (4)}}{6}$

Step 2 - $x = \frac{8 \pm \sqrt{- 64 - 48}}{6}$

Here is how to find the error:

First, Zariah substituted $a = 3$ , $b = - 8$ , and $c = 4$ into the formula in Step 1.

However, under the radical, $b^{2} = (- 8)^{2} = + 64$ .

Step 2 would then become:

$x = \frac{8 \pm \sqrt{64 - 48}}{6}$

Which leads to the following solution:

$x = \frac{8 \pm \sqrt{16}}{6}$

$x = \frac{8 \pm 4}{6}$

$x = \frac{12}{6} = 2$ , $x = \frac{4}{6} = \frac{2}{3}$