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

Sagot :

Xignersoidn

[tex]−5(²+2−4)[/tex]

[tex]−5²−10+20[/tex]

Solution :—

[tex]−5²−10+20[/tex]

Devishri1977idn

Answer:

Step-by-step 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

[tex]\sqrt{D}=\sqrt{2*2*3*5*7}=2\sqrt{105}\\\\x=\dfrac{-b+\sqrt{D}}{2a} \ ; x =\dfrac{-b-\sqrt{D}}{2a}\\\\\\x=\dfrac{10+2\sqrt{105}}{2*(-5)} \ ; \ x =\dfrac{10-2\sqrt{105}}{2*(-5)}\\\\\\x = \dfrac{2*(5+\sqrt{105})}{2*(-5)} ; \ x =\dfrac{2*(5-\sqrt{D})}{2*(-5)}\\\\\\x =\dfrac{5+\sqrt{105}}{-5} \ ; \ x=\dfrac{5-\sqrt{105}}{-5}\\\\\\x=\dfrac{-5-\sqrt{105}}{5} \ ; \ x =\dfrac{-5+\sqrt{105}}{5}[/tex]

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