Someone please help me with problem #7 !!!
1 answer:
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Answer:
see explanation
Step-by-step explanation:
(a)
x² - 36 = 0 ← is a difference of squares and factors as
(x - 6)(x + 6) = 0
equate each factor to zero and solve for x
x + 6 = 0 ⇒ x = - 6
x - 6 = 0 ⇒ x = 6
(b)
x² - 5x + 4 = 0
consider the product of the factors of the constant term (+ 4) which sum to give the coefficient of the x- term (- 5)
the factors are - 1 and - 4 , since
- 1 × - 4 = + 4 and - 1 - 4 = - 5 , then
(x - 1)(x - 4) = 0 ← in factored form
equate each factor to zero and solve for x
x - 1 = 0 ⇒ x = 1
x - 4 = 0 ⇒ x = 4
(c)
x² - 2x = 3 ( subtract 3 from both sides )
x² - 2x - 3 = 0 ← in standard form
consider the product of the factors of the constant term (- 3) which sum to give the coefficient of the x- term (- 2)
the factors are + 1 and - 3 , since
1 × - 3 = - 3 and 1 - 3 = - 2 , then
(x + 1)(x - 3) = 0 ← in factored form
equate each factor to zero and solve for x
x + 1 = 0 ⇒ x = - 1
x - 3 = 0 ⇒ x = 3
(d)
6x² - 11x - 10 = 0
consider the product of the factors of the coefficient of the x² term and the constant term which sum to give the coefficient of the x- term.
product = 6 × - 10 = - 60 and sum = - 11
the factors are + 4 and - 15
use these factors to split the x- term
6x² + 4x - 15x - 10 = 0 ( factor the first/second and third/fourth terms )
2x(3x + 2) - 5(3x + 2) = 0 ← factor out (3x + 2) from each term
(3x + 2)(2x - 5) = 0
equate each factor to zero and solve for x
3x + 2 = 0 ⇒ 3x = - 2 ⇒ x = -
2x - 5 = 0 ⇒ 2x = 5 ⇒ x =