Question Find the values of $a$ that satisfy the equation $15(2a-2) = 5(a^2-1)$ .

Studdy Solution

STEP 1

Assumptions
1. We are given the equation $15(2a - 2) = 5(a^2 - 1)$ .
2. We need to solve for the variable $a$ .

STEP 2

First, we need to simplify both sides of the equation. Let's start by distributing the multiplication on the left side.
$15 \times (2a - 2) = 5 \times (a^2 - 1)$

STEP 3

Carry out the multiplication on the left side.
$30a - 30 = 5 \times (a^2 - 1)$

STEP 4

Now distribute the multiplication on the right side.
$30a - 30 = 5a^2 - 5$

STEP 5

To solve the quadratic equation, we need to bring all terms to one side to set the equation to zero.
$5a^2 - 30a + 25 = 0$

STEP 6

The quadratic equation is now in the form $5a^2 - 30a + 25 = 0$ . We can try to factor this equation, or use the quadratic formula to find the solutions for $a$ . Let's first check if the equation is factorable.

STEP 7

We are looking for two numbers that multiply to $5 \times 25 = 125$ and add up to $-30$ . After checking possible factors, we find that the numbers $-5$ and $-25$ satisfy these conditions.

STEP 8

Now we can factor the quadratic equation using these numbers.
$5a^2 - 5a - 25a + 25 = 0$

STEP 9

Factor by grouping. Group the first two terms and the last two terms.
$(5a^2 - 5a) - (25a - 25) = 0$

STEP 10

Factor out the common factor in each group.
$5a(a - 1) - 25(a - 1) = 0$

STEP 11

Now, factor out the common binomial factor $(a - 1)$ .
$(a - 1)(5a - 25) = 0$

STEP 12

Further simplify the second factor by factoring out the common factor of 5.
$(a - 1)(5(a - 5)) = 0$

STEP 13

Apply the zero-product property, which states that if a product of two factors is zero, then at least one of the factors must be zero.
$a - 1 = 0 \quad \text{or} \quad 5(a - 5) = 0$

STEP 14

Solve each equation for $a$ .
First equation:
$a - 1 = 0$
$a = 1$
Second equation:
$5(a - 5) = 0$
$a - 5 = 0$
$a = 5$

STEP 15

The solutions to the equation $15(2a - 2) = 5(a^2 - 1)$ are $a = 1$ and $a = 5$ .
Therefore, the correct answer is $a=5, a=1$ .

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