Question

Solve the following quadratic equations


`x { }^(2) - x - 42`

`x {}^(2) + x - 6`

`r {}^(2) - 25`
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Answer 1

see explanation

Explanation:

To solve these equations equate them to zero, that is

x² - x - 42 = 0

Consider the factors of the constant term (- 42) which sum to give the coefficient of the x- term (- 1)

The factors are - 7 and + 6, since

- 7 × 6 = - 42 and - 7 + 6 = - 1 , thus

(x - 7)(x + 6) = 0

Equate each factor to zero and solve for x

x - 7 = 0 ⇒ x = 7

x + 6 = 0 ⇒ x = - 6

Solutions are x = - 6, x = 7

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x² + x - 6 = 0 ← in standard form

(x + 3)(x - 2) = 0 ← in factored form

Equate each factor to zero and solve for x

x + 3 = 0 ⇒ x = - 3

x - 2 = 0 ⇒ x = 2

Solutions are x = - 3, x = 2

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x² - 25 = 0 ( add 25 to both sides )

x² = 25 ( take the square root of both sides )

x = ±
`√(25)` = ± 5

Solutions are x = - 5, x = 5