Mathematics, 02.02.2020 22:03 minecraft56

What are the roots of the polynomial equation x3-7x=6x-12?

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This is not a polynomial equation unless one of those is squared. As it stands x=-.833. If you can tell me which is squared I can help solve the polynomial.

Ok, that is usually notated as x^3 to be clear. I'll solve it now.

x^3-13x-12=0

Then use factor theorum to solve x^3-13x-12/x+1 =0
So you get one solution of x+1=0
x=-1

Then you have x^2-x-12 now you complete the square.

Take half of the x-term coefficient and square it. Add this value to both sides. In this example we have:

The x-term coefficient = −1

The half of the x-term coefficient = −1/2

After squaring we have (−1/2)2=1/4

When we add 1/4 to both sides we have:

x2−x+1/4=12+1/4

STEP 3: Simplify right side

x2−x+1/4=49/4

STEP 4: Write the perfect square on the left.

(x−1/2)2=49/4

STEP 5: Take the square root of both sides.

x−1/2=±√49/4

STEP 6: Solve for x.

x=1/2±√49/4

that is,

x1=−3

x2=4

and the one from before

x=-1

Answer from: harleycochran2ovyt3n

This is not a polynomial equation unless one of those is squared.

Answer from: Quest

1 1/4 hours

step-by-step explanation:

2/4 + 2/4 + 1/4 = 5/4 or 1 1/4

Answer from: Quest

the correct functions are

h(x) = 3x + 2

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values of k (x)

case 1) k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option 4. k(x) = 2^(3x + 2 )

case 2) k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option 1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) = [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option 3.) k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) = [2^x]-[3x + 2]=2^x-3x-> is the option 2.) k(x) = 2^x - 3x - 2

case 5) k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option 5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option 6. k(x) = 3(2^x) +2

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