Exercise 38 Page 418

We will solve the given equation by taking the square roots. Remember to consider the positive and negative solutions.

r^{2} - 3 = 61

$LHS + 3 = RHS + 3$

r^{2} = 64

$LHS = RHS$

r^{2} = 64

$a^{2} = \pm a$

r = \pm 64

Calculate root

r = \pm 8

We found that

r = \pm 8 .

Thus, there are two solutions for the equation, which are

r = 8

and

r = - 8 .

Checking our answer

We can check our answers by substituting them for

r

in the given equation. Let's start with

r = - 8 .

r^{2} - 3 = 61

$r = - 8$

(- 8)^{2} - 3 = ? 61

▼

Simplify

$(- a)^{2} = a^{2}$

8^{2} - 3 = ? 61

Calculate power

64 - 3 = ? 61

Subtract term

61 = 61 ✓

Since

61 = 61,

we know that

r = - 8

is a solution of the equation. Let's check if

r = 8

is also a solution.

r^{2} - 3 = 61

$r = 8$

8^{2} - 3 = ? 61

▼

Simplify

Calculate power

64 - 3 = ? 61

Subtract term

61 = 61 ✓

Again, since

61 = 61,

we know that

r = 8

is a solution of the equation.