Question

Solve the equations by factoring:x^2−3x−40=04x^2−81=0

Answer 1

Answer:

see explanation

Step-by-step explanation:

Given

x² - 3x - 40 = 0

Consider the factors of the constant term (- 40) which sum to give the coefficient of the x- term (- 3)

The factors are - 8 and + 5, since

- 8 × 5 = - 40 and - 8 + 5 = - 3, thus

(x - 8)(x + 5) = 0 ← in factored form

Equate each factor to zero and solve for x

x - 8 = 0 ⇒ x = 8

x + 5 = 0 ⇒ x = 5

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Given

4x² - 81 = 0 ← this is a difference of squares and factors in general as

a² - b² = (a + b)(a - b), thus

4x² - 81

=(2x)² - 9²

= (2x + 9)(2x - 9) = 0 ← in factored form

Equate each factor to zero and solve for x

2x + 9 = 0 ⇒ 2x = - 9 ⇒ x = - $\frac{9}{2}$

2x - 9 = 0 ⇒ 2x = 9 ⇒ x = $\frac{9}{2}$

Answer 2

Answer:  (a) x = 8, x = -5

                $\bold{(b)\ x=\pm \dfrac{9}{2}}$

Step-by-step explanation:

(a) Find two numbers that multiply to get -40 and add to get -3.

x² - 3x - 40 = 0

              ∧

            1  -40

            2  -20

            4  -10

            5   -8  = -3

So 5 and -8 go into the parenthesis:

(x + 5) (x - 8) = 0

Set each factor equal to zero and solve for x:

x + 5 = 0     x - 8 = 0

  x = -5           x = 8

(b) Since the middle term is missing, isolate x² and then square root both sides.

$4x^2-81=0\\\\\\4x^2=81\\\\\\x^2=\dfrac{81}{4}\\\\\\\sqrt{x^2}=\sqrt{\dfrac{81}{4}}\\\\\\\\x=\pm\dfrac{9}{2}$