Question Find the values of $x$ where the equations $y = x^2 - 24$ and $y = x - 12$ intersect.

Studdy Solution

STEP 1

Assumptions
1. We have two equations: $y = x^2 - 24$ $y = x - 12$
2. We need to find the values of $x$ and $y$ that satisfy both equations simultaneously.

STEP 2

Since both equations are equal to $y$ , we can set them equal to each other to find the value of $x$ .
$x^2 - 24 = x - 12$

STEP 3

To solve for $x$ , we need to rearrange the equation to form a quadratic equation. We do this by moving all terms to one side of the equation.
$x^2 - x - 24 + 12 = 0$

STEP 4

Simplify the equation by combining like terms.
$x^2 - x - 12 = 0$

STEP 5

Now we have a quadratic equation in standard form. We can solve for $x$ by factoring, completing the square, or using the quadratic formula. Let's try to factor the equation first.
We are looking for two numbers that multiply to $-12$ and add up to $-1$ .

STEP 6

After examining possible factors of $-12$ , we find that $-4$ and $3$ are the numbers that satisfy the conditions.
So we can write the factored form of the quadratic equation as:
$(x - 4)(x + 3) = 0$

STEP 7

Now we can use the zero product property, which states that if the product of two factors is zero, then at least one of the factors must be zero.
So we set each factor equal to zero:
$x - 4 = 0 \quad \text{or} \quad x + 3 = 0$

STEP 8

Solve each equation for $x$ .
For $x - 4 = 0$ : $x = 4$
For $x + 3 = 0$ : $x = -3$

STEP 9

Now that we have the values of $x$ , we can substitute them into either of the original equations to find the corresponding values of $y$ .
Let's use the second equation $y = x - 12$ because it is simpler.
First, substitute $x = 4$ :
$y = 4 - 12$

STEP 10

Calculate the value of $y$ when $x = 4$ .
$y = -8$

STEP 11

Now, substitute $x = -3$ into the second equation:
$y = -3 - 12$

STEP 12

Calculate the value of $y$ when $x = -3$ .
$y = -15$

STEP 13

We have found two solutions to the system of equations:
When $x = 4$ , $y = -8$ . When $x = -3$ , $y = -15$ .
Thus, the solutions to the system of equations are $(4, -8)$ and $(-3, -15)$ .

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