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Question 13 (a) Solve the equation $x^2 + 7x + 12 = 0$ - Junior Cycle Mathematics - Question 13 - 2015

Question 13

Question 13 (a) Solve the equation $x^2 + 7x + 12 = 0$. (b) Verify one of your answers from (a).

Worked Solution & Example Answer:Question 13 (a) Solve the equation $x^2 + 7x + 12 = 0$ - Junior Cycle Mathematics - Question 13 - 2015

Step 1

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Answer

To solve the quadratic equation, we apply the quadratic formula:

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

In this equation, we identify $a = 1$ , $b = 7$ , and $c = 12$ .
Substituting these values into the formula, we have:

$x = \frac{-7 \pm \sqrt{7^2 - 4(1)(12)}}{2(1)}$
$= \frac{-7 \pm \sqrt{49 - 48}}{2}$
$= \frac{-7 \pm \sqrt{1}}{2}$

This simplifies to: $x = \frac{-7 - 1}{2} = -4$
and
$x = \frac{-7 + 1}{2} = -3$

Thus, the solutions to the equation are $x = -4$ and $x = -3$ .

Step 2

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Answer

Let's verify $x = -3$ . We will substitute $x = -3$ back into the original quadratic equation:

$(-3)^2 + 7(-3) + 12 = 0$

Calculating each term, we get: $9 - 21 + 12 = 0$

Which simplifies to: $9 - 21 + 12 = 0,$ confirming that this equation holds true.

Thus, $x = -3$ is indeed a valid solution.

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