Solve: [tex]16x^2 - 80 = 0[/tex]

Round the answer to the nearest hundredth.

A. [tex]x = -0.45[/tex] and [tex]x = 0.45[/tex]
B. [tex]x = -2.24[/tex] and [tex]x = 2.24[/tex]
C. [tex]x = -8.94[/tex] and [tex]x = 8.94[/tex]
D. [tex]x = -9.80[/tex] and [tex]x = 9.80[/tex]



Answer :

To solve the quadratic equation [tex]\(16x^2 - 80 = 0\)[/tex] and round the answers to the nearest hundredth, follow these detailed steps:

1. Identify the standard form and coefficients:
The given equation is [tex]\(16x^2 - 80 = 0\)[/tex].
This is a quadratic equation, which generally has the form [tex]\(ax^2 + bx + c = 0\)[/tex].
By comparing, we get:
[tex]\[ a = 16, \quad b = 0, \quad c = -80 \][/tex]

2. Calculate the discriminant:
The discriminant [tex]\(\Delta\)[/tex] for a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[ \Delta = 0^2 - 4 \times 16 \times (-80) \][/tex]
Simplifying:
[tex]\[ \Delta = 0 + 5120 = 5120 \][/tex]

3. Find the roots using the quadratic formula:
The quadratic formula is:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
Substituting the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(\Delta\)[/tex]:
[tex]\[ x = \frac{-0 \pm \sqrt{5120}}{2 \times 16} \][/tex]
Simplifying inside the square root:
[tex]\[ x = \frac{\pm \sqrt{5120}}{32} \][/tex]
Calculating [tex]\(\sqrt{5120}\)[/tex]:
[tex]\[ \sqrt{5120} \approx 71.55417527999327 \][/tex]
Therefore:
[tex]\[ x = \frac{71.55417527999327}{32} \quad \text{and} \quad x = \frac{-71.55417527999327}{32} \][/tex]
Simplifying both expressions:
[tex]\[ x \approx 2.23606797749979 \quad \text{and} \quad x \approx -2.23606797749979 \][/tex]

4. Round the answers to the nearest hundredth:
[tex]\[ x \approx 2.24 \quad \text{and} \quad x \approx -2.24 \][/tex]

Therefore, the solutions to the equation [tex]\(16x^2 - 80 = 0\)[/tex] rounded to the nearest hundredth are:
[tex]\[ x = 2.24 \quad \text{and} \quad x = -2.24 \][/tex]

The correct answer from the given choices is:
[tex]\[ \boxed{x = -2.24 \quad \text{and} \quad x = 2.24} \][/tex]

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