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What are the solutions of this quadratic equation?

[tex]\[ 2x^2 + 16x - 8 = 0 \][/tex]

A. [tex]\( x = -4 \pm 2 \sqrt{5} \)[/tex]

B. [tex]\( x = 4 \pm 2 \sqrt{5} \)[/tex]

C. [tex]\( x = -2 \pm 4 \sqrt{5} \)[/tex]

D. [tex]\( x = 2 \# 4 \sqrt{5} \)[/tex]


Sagot :

To solve the quadratic equation [tex]\(2x^2 + 16x - 8 = 0\)[/tex], we need to use the quadratic formula:

[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

Here, [tex]\(a = 2\)[/tex], [tex]\(b = 16\)[/tex], and [tex]\(c = -8\)[/tex].

First, let's calculate the discriminant:

[tex]\[ \Delta = b^2 - 4ac \][/tex]

Substituting the values, we get:

[tex]\[ \Delta = 16^2 - 4 \times 2 \times (-8) \][/tex]

[tex]\[ \Delta = 256 + 64 \][/tex]

[tex]\[ \Delta = 320 \][/tex]

Now, we find the square root of the discriminant:

[tex]\[ \sqrt{\Delta} = \sqrt{320} = \sqrt{64 \times 5} = 8\sqrt{5} \][/tex]

Next, apply the quadratic formula:

[tex]\[ x = \frac{-16 \pm 8\sqrt{5}}{2 \times 2} \][/tex]

Simplify the expression:

[tex]\[ x = \frac{-16 \pm 8\sqrt{5}}{4} \][/tex]

[tex]\[ x = \frac{-16}{4} \pm \frac{8\sqrt{5}}{4} \][/tex]

[tex]\[ x = -4 \pm 2\sqrt{5} \][/tex]

Thus, the solutions to the quadratic equation [tex]\(2x^2 + 16x - 8 = 0\)[/tex] are:

[tex]\[ x = -4 + 2\sqrt{5} \quad \text{and} \quad x = -4 - 2\sqrt{5} \][/tex]

The correct answer is:

A. [tex]\(x = -4 \pm 2\sqrt{5}\)[/tex]