One-third times the difference of a number and 5 is Negative two-thirds. Which equation best shows this? One possible first step
2 answers:
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Answer:
see explanation
Step-by-step explanation:
Given
4 - 5a² + 1 = 0
Use the substitution u = a², then equation is
4u² - 5u + 1 = 0
Consider the product of the coefficient of the u² term and the constant term
product = 4 × 1 = 4 and sum = - 5
The factors are - 4 and - 1
Use these factors to split the u- term
4u² - 4u - u + 1 = 0 ( factor the first/second and third/fourth terms )
4u(u - 1) - 1(u - 1) = 0 ← factor out (u - 1) from each term
(u - 1)(4u - 1) = 0
Equate each factor to zero and solve for u
u - 1 = 0 ⇒ u = 1
4u - 1 = 0 ⇒ 4u = 1 ⇒ u =
Convert u back into terms of a, that is
a² = 1 ⇒ a = ± 1
a² = ⇒ a = ±
Solutions are a = ± 1 , a = ±