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Kobotan [32]
3 years ago
15

​Find all roots: x^3 + 7x^2 + 12x = 0 Show all work and check your answer.

Mathematics
1 answer:
Aliun [14]3 years ago
4 0

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

x\left(x^{2}+7 x+12\right)=0   ----- eqn 1

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:

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

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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Write an equation perpendicular to x - 4y = 20 that passes through the point (4, -3)
Nina [5.8K]

Answer:

y = - 4x + 13

Step-by-step explanation:

In order to do this problem you would need a whole new equation, but you will use what is given to you:

instead of going with:

x - 4y = 20

Turn it into y = mx + b form, also known as slope-intercept form:

x - 4y = 20

-x       = -x

_________

- 4y = 20 - x

Then divide both sides by - 4:

\frac{-4y}{-4} = \frac{-x + 20}{-4}

You will then get:

y = \frac{1}{4}x - 5

From this equation we will only need the slope!

m || = \frac{1}{4}

(This slope is for parallel)

but if you want perpendicular to the slope, then we need the negative reciprocal!

Negative reciprocal is a flipped version of the value that is negative, for example:

2 = -\frac{1}{2}

Because \frac{2}{1} = 2

Now we will find the perpendicular slope which is:

m ⊥ = - 4

Now substitute this slope into slope intercept form:

y = mx + b

(-3) = -4(4) + b

(Take away parentheses)

-3 = -16 + b

(Move - 3 to the other side, by making it positive, but what you do to one side, you do to the other)

= -13 + 6

(Move - 13 to the other side, by make - 13 into +13)

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Your answer is:

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8 0
3 years ago
11111111111111111111111
Nat2105 [25]

Answer:

1) Zero based on (-16·t - 2) is t = -1/8 second

2) Zero based on (t - 1) is t = 1 second

Step-by-step explanation:

The given functions representing the height of the beach ball the child throws as a function of time are;

y = (-16·t - 2)·(t - 1) and y = -16·t² + 14·t + 2

We note that (-16·t - 2)·(t - 1) = -16·t² + 14·t + 2

Therefore, the function representing the height of the beachball, 'y', is y = (-16·t - 2)·(t - 1) = -16·t² + 14·t + 2

The zeros of a function are the values of the variables, 'x', of the function that makes the value of the function, f(x), equal to zero

In the function of the question, we have;

y = (-16·t - 2)·(t - 1) = -16·t² + 14·t + 2

The above equation can be written as follows;

y = (-16·t - 2) × (t - 1)

Therefore, 'y' equals zero when either (-16·t - 2) = 0 or (t - 1) = 0

1) The zero based on (-16·t - 2) = 0, is given as follows;

(-16·t - 2) = 0

∴ t = 2/(-16) = -1/8

t = -1/8 second

The zero based on (-16·t - 2) is t = -1/8 second

2) The zero based on (t - 1) = 0, is given as follows;

(t - 1) = 0

∴ t = 1 second

The zero based on (t - 1) is t = 1 second

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