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3 years ago

13

26 ptss!! select the expressions that are equivalent to the expression below.

1 answer:

valkas [14]3 years ago

7 0

Answer:

1st, 2nd, and last.

Step-by-step explanation:

For the first two common rules can be applied such as $n\sqrt{a/b}$ = $\frac{\sqrt[n]{a} }{\sqrt[n]{b} }$ and that $\sqrt[n]{\frac{a}{b} } = (\frac{a}{b})^{n}$

and for the last one this is just simplifying the radical and if you factor it you realize that, $\frac{750}{512} = \frac{5^3}{8^3}*6$ meaning that you can take the radical off of 5/8 and end up with that final answer

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Aliun [14]

The three roots of x^3 + 7x^2 + 12x = 0 is 0,-3 and -4

<u>Solution:</u>

We have been given a cubic polynomial.

$x^{3}+7 x^{2}+12 x=0$

We need to find the three roots of the given polynomial.

Since it is a cubic polynomial, we can start by taking ‘x’ common from the equation.

This gives us:

$x^{3}+7 x^{2}+12 x=0$

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So, from the above eq1 we can find the first root of the polynomial, which will be:

x = 0

Now, we need to find the remaining two roots which are taken from the remaining part of the equation which is:

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we have to use the quadratic equation to solve this polynomial. The quadratic formula is:

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

Now, a = 1, b = 7 and c = 12

By substituting the values of a,b and c in the quadratic equation we get;

$\begin{array}{l}{x=\frac{-7 \pm \sqrt{7^{2}-4 \times 1 \times 12}}{2 \times 1}} \\\\{x=\frac{-7 \pm \sqrt{1}}{2}}\end{array}$

<em><u>Therefore, the two roots are:</u></em>

$\begin{array}{l}{x=\frac{-7+\sqrt{1}}{2}=\frac{-7+1}{2}=\frac{-6}{2}} \\\\ {x=-3}\end{array}$

And,

$\begin{array}{c}{x=\frac{-7-\sqrt{1}}{2}} \\\\ {x=-4}\end{array}$

Hence, the three roots of the given cubic polynomial is 0, -3 and -4

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