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What are the roots of the quadratic equation 2x2+7x+4=0? Select all that apply.

A. −7+17√4
B. −7−17√4
C. −7+17√2
D.−7−17√2
E. 7+17√4
F. 7−17√4
G. 7+17√2
H. 7−17√2

1 Answer

2 votes

Answer:

option A and B


x=\frac{-7+√(17)} {4}

and


x=\frac{-7-√(17)} {4}

Explanation:

we have


2x^2+7x+4=0

The formula to solve a quadratic equation of the form


ax^(2) +bx+c=0

is equal to


x=\frac{-b\pm\sqrt{b^(2)-4ac}} {2a}

in this problem we have


2x^2+7x+4=0

so


a=2\\b=7\\c=4

substitute in the formula


x=\frac{-7\pm\sqrt{7^(2)-4(2)(4)}} {2(2)}


x=\frac{-7\pm√(17)} {4}

so


x=\frac{-7+√(17)} {4}

and


x=\frac{-7-√(17)} {4}

User Ain Tohvri
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