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3 years ago

Twelve less than the cube of the sum of six times a number and three.

Mathematics

1 answer:

makvit [3.9K]3 years ago

8 0

Answer:

$(6 * n + 3)^{3}$ - 12

Step-by-step explanation:

Twelve less than the cube of the sum of six times a number and three.

1. Change the words into numbers and math symbols.

Twelve = 12

Less than = subtract -

sum of six times a number and three = (6 * n + 3)

cube = exponent 3

2. Write the problem.

$(6 * n + 3)^{3}$ - 12

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<h3>Solution:</h3>

Given quadratic equation is $2 x^{2}+3 x-8=0$

The general quadratic equation is of form:

$a x^{2}+b x+c=0$

Now comparing the general equation with the given equation we get

a = 2 , b = 3 and c = -8

The formula to determine roots of the quadratic equation is:

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

On plugging in vlaues, we get

$x=\frac{-3 \pm \sqrt{3^{2}-4 \times 2 \times(-8)}}{2 \times 2}$

$\mathrm{x}=\frac{-3 \pm \sqrt{9-(-64)}}{4}$

On solving we get,

$\begin{array}{l}{\mathrm{x}=\frac{-3 \pm \sqrt{9+64}}{4}} \\\\ {\mathrm{x}=\frac{-3 \pm \sqrt{73}}{4}}\end{array}$

$\mathrm{x}=\frac{-3+\sqrt{73}}{4} \text { OR } \mathrm{x}=\frac{-3-\sqrt{73}}{4}$

x = 1.39 OR x = -2.89

Hence , the positive solution to the quadratic equation is x = 1.39

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