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Find all real zeros of the function algebraically -5(x^2+2x-4)

User Vernell
by
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2 Answers

9 votes


−5(²+2−4)


−5²−10+20

Solution :—


−5²−10+20

User Soup In Boots
by
8.6k points
12 votes

Answer:

Explanation:

-5(x² + 2x - 4) = -5x² - 10x + 20

a = -4 ; b = -10 ; c = 20

D = b² - 4ac = (-10)² - 4*(-4)*20

= 100 + 320 = 420


√(D)=√(2*2*3*5*7)=2√(105)\\\\x=(-b+√(D))/(2a) \ ; x =(-b-√(D))/(2a)\\\\\\x=(10+2√(105))/(2*(-5)) \ ; \ x =(10-2√(105))/(2*(-5))\\\\\\x = (2*(5+√(105)))/(2*(-5)) ; \ x =(2*(5-√(D)))/(2*(-5))\\\\\\x =(5+√(105))/(-5) \ ; \ x=(5-√(105))/(-5)\\\\\\x=(-5-√(105))/(5) \ ; \ x =(-5+√(105))/(5)

User Dan Billings
by
8.4k points

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