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Artyom0805 [142]

3 years ago

5

What are the x-intercepts of 4x^ + 8x - 5

1 answer:

mojhsa [17]3 years ago

6 0

Assuming the exponent is supposed to be "^2" your equation will read:

4x² + 8x - 5

It must set equal to y to be a valid function and the y must be set equal to zero to find x-intercepts.

4x² + 8x - 5 = 0
4x² - 2x + 10x - 5 = 0
2x(2x - 1) + 5(2x - 1) = 0

(2x + 5)(2x - 1) = 0

Set each binomial equal to zero.

2x + 5 = 0
2x = 0 - 5
2x = - 5
Divide both sides by 2
x = - 5/2

2x - 1 = 0
2x = 0 + 1
2x = 1
x = 1/2

Your x-intercepts are x = - 5/2, 1/2 or (- 5/2, 0) and (1/2, 0)

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Given:

$\angle A=5 x-18^{\circ}$

$\angle B=3 x+42^{\circ}$

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Alternate interior angle theorem:

If two parallel lines cut by a transversal, then the alternate interior angles are congruent.

⇒ m∠A = m∠B

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