Polynomial equation, cannot solve for $x$
$$3x^2-4x-4+x^3=x^3+2x+2$$
This boils down to (I think):
$$3x^2 - 6x - 6 = 0$$
I'm trying to solve for $x$ using the polynomial equation:
$$begin{align}
x &= frac{-bpmsqrt{b^2-4ac}}{2a} \
a &= 3 \[0.2ex]
b &= -6 \
c &= -6
end{align}$$
My textbook tells me that the solutions for the positive and negative versions of $x$ are $x = 1 + sqrt{3}$ and $x = 1 - sqrt{3}$. I'm unable to replicate this.
Here's my work:
$$begin{align}
x&=frac{-bpmsqrt{b^2-4ac}}{2a} \[0.8ex]
x&=frac{-6pmsqrt{(-6)^2-4(3)(-6)}}{2a} \[0.8ex]
x&=frac{-6pmsqrt{36-72}}{2a}
end{align}$$
Since $(36 - 72) < 0$, I'm unable to take the square root so I do not see how the equation is solvable.
Unless I boiled down the equation into the wrong form with the original.
I turned:
$$3x^2-4x-4+x^3=x^3+2x+2$$
into
$$3x^2-6x-6=0$$
How can I arrive at solutions for $x$ where $x = 1 + sqrt{3}$ and $x = 1 - sqrt{3},$ per my textbook's given solutions?
polynomials
edited Nov 20 '18 at 21:01
Robert Howard
1,9161822
asked Sep 17 '18 at 15:16
Doug Fir
1837
add a comment |
$$3x^2-4x-4+x^3=x^3+2x+2$$
This boils down to (I think):
$$3x^2 - 6x - 6 = 0$$
I'm trying to solve for $x$ using the polynomial equation:
$$begin{align}
x &= frac{-bpmsqrt{b^2-4ac}}{2a} \
a &= 3 \[0.2ex]
b &= -6 \
c &= -6
end{align}$$
My textbook tells me that the solutions for the positive and negative versions of $x$ are $x = 1 + sqrt{3}$ and $x = 1 - sqrt{3}$. I'm unable to replicate this.
Here's my work:
$$begin{align}
x&=frac{-bpmsqrt{b^2-4ac}}{2a} \[0.8ex]
x&=frac{-6pmsqrt{(-6)^2-4(3)(-6)}}{2a} \[0.8ex]
x&=frac{-6pmsqrt{36-72}}{2a}
end{align}$$
Since $(36 - 72) < 0$, I'm unable to take the square root so I do not see how the equation is solvable.
Unless I boiled down the equation into the wrong form with the original.
I turned:
$$3x^2-4x-4+x^3=x^3+2x+2$$
into
$$3x^2-6x-6=0$$
How can I arrive at solutions for $x$ where $x = 1 + sqrt{3}$ and $x = 1 - sqrt{3},$ per my textbook's given solutions?
polynomials
edited Nov 20 '18 at 21:01
Robert Howard
1,9161822
asked Sep 17 '18 at 15:16
Doug Fir
1837
1
discriminant will be $36+72$, that's your mistake
– Vasya
Sep 17 '18 at 15:34
add a comment |
$$3x^2-4x-4+x^3=x^3+2x+2$$
This boils down to (I think):
$$3x^2 - 6x - 6 = 0$$
I'm trying to solve for $x$ using the polynomial equation:
$$begin{align}
x &= frac{-bpmsqrt{b^2-4ac}}{2a} \
a &= 3 \[0.2ex]
b &= -6 \
c &= -6
end{align}$$
My textbook tells me that the solutions for the positive and negative versions of $x$ are $x = 1 + sqrt{3}$ and $x = 1 - sqrt{3}$. I'm unable to replicate this.
Here's my work:
$$begin{align}
x&=frac{-bpmsqrt{b^2-4ac}}{2a} \[0.8ex]
x&=frac{-6pmsqrt{(-6)^2-4(3)(-6)}}{2a} \[0.8ex]
x&=frac{-6pmsqrt{36-72}}{2a}
end{align}$$
Since $(36 - 72) < 0$, I'm unable to take the square root so I do not see how the equation is solvable.
Unless I boiled down the equation into the wrong form with the original.
I turned:
$$3x^2-4x-4+x^3=x^3+2x+2$$
into
$$3x^2-6x-6=0$$
How can I arrive at solutions for $x$ where $x = 1 + sqrt{3}$ and $x = 1 - sqrt{3},$ per my textbook's given solutions?
polynomials
edited Nov 20 '18 at 21:01
Robert Howard
1,9161822
asked Sep 17 '18 at 15:16
Doug Fir
1837
$$3x^2-4x-4+x^3=x^3+2x+2$$
This boils down to (I think):
$$3x^2 - 6x - 6 = 0$$
I'm trying to solve for $x$ using the polynomial equation:
$$begin{align}
x &= frac{-bpmsqrt{b^2-4ac}}{2a} \
a &= 3 \[0.2ex]
b &= -6 \
c &= -6
end{align}$$
My textbook tells me that the solutions for the positive and negative versions of $x$ are $x = 1 + sqrt{3}$ and $x = 1 - sqrt{3}$. I'm unable to replicate this.
Here's my work:
$$begin{align}
x&=frac{-bpmsqrt{b^2-4ac}}{2a} \[0.8ex]
x&=frac{-6pmsqrt{(-6)^2-4(3)(-6)}}{2a} \[0.8ex]
x&=frac{-6pmsqrt{36-72}}{2a}
end{align}$$
Since $(36 - 72) < 0$, I'm unable to take the square root so I do not see how the equation is solvable.
Unless I boiled down the equation into the wrong form with the original.
I turned:
$$3x^2-4x-4+x^3=x^3+2x+2$$
into
$$3x^2-6x-6=0$$
How can I arrive at solutions for $x$ where $x = 1 + sqrt{3}$ and $x = 1 - sqrt{3},$ per my textbook's given solutions?
polynomials
polynomials
edited Nov 20 '18 at 21:01
Robert Howard
1,9161822
asked Sep 17 '18 at 15:16
Doug Fir
1837
edited Nov 20 '18 at 21:01
Robert Howard
1,9161822
asked Sep 17 '18 at 15:16
Doug Fir
1837
edited Nov 20 '18 at 21:01
Robert Howard
1,9161822
edited Nov 20 '18 at 21:01
Robert Howard
1,9161822
edited Nov 20 '18 at 21:01
Robert Howard
1,9161822
1,9161822
asked Sep 17 '18 at 15:16
Doug Fir
1837
asked Sep 17 '18 at 15:16
Doug Fir
1837
asked Sep 17 '18 at 15:16
Doug Fir
1837
1837
1
discriminant will be $36+72$, that's your mistake
– Vasya
Sep 17 '18 at 15:34
add a comment |
1
discriminant will be $36+72$, that's your mistake
– Vasya
Sep 17 '18 at 15:34
1
1
discriminant will be $36+72$, that's your mistake
– Vasya
Sep 17 '18 at 15:34
discriminant will be $36+72$, that's your mistake
– Vasya
Sep 17 '18 at 15:34
add a comment |
2 Answers
2
active
oldest
votes
so we start off with:
$$x^3+3x^2-4x-4=x^3+2x+2$$
which we can re-write as:
$$3x^2-6x-6=0$$
so we know that:
$$x=frac{-bpmsqrt{b^2-4ac}}{2a}=frac{6pmsqrt{36-(4)(3)(-6)}}{6}=1pmfrac{sqrt{108}}{6}=1pmsqrt{3}$$
The mistake you made was $36-72$ instead of $36+72$
answered Sep 17 '18 at 15:38
Henry Lee
1,713218
Thanks but would 108 not be 114 since the initial 6 +-?
– Doug Fir
Sep 17 '18 at 15:49
1
that is not inside the squareroot so we do not include it
– Henry Lee
Sep 17 '18 at 15:51
mathwarehouse.com/dictionary/D-words/… remember where you can apply your commutative and distributive laws
– Henry Lee
Sep 17 '18 at 15:52
add a comment |
$$x^3+3x^2-4x-4=x^3+2x+2$$
$$3x^2-6x-6=0$$
$$x^2-2x-2=0$$
$$x=frac{-(-2)pmsqrt{(-2)^2-4(1)(-2)}}{2(1)}$$
$$x=frac{2pmsqrt{4+8}}{2}=frac{2pm 2sqrt 3}{2}=1pmsqrt 3$$
answered Sep 17 '18 at 15:44
Deepesh Meena
4,29921025
add a comment |
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2 Answers
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2 Answers
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active
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active
oldest
votes
so we start off with:
$$x^3+3x^2-4x-4=x^3+2x+2$$
which we can re-write as:
$$3x^2-6x-6=0$$
so we know that:
$$x=frac{-bpmsqrt{b^2-4ac}}{2a}=frac{6pmsqrt{36-(4)(3)(-6)}}{6}=1pmfrac{sqrt{108}}{6}=1pmsqrt{3}$$
The mistake you made was $36-72$ instead of $36+72$
answered Sep 17 '18 at 15:38
Henry Lee
1,713218
Thanks but would 108 not be 114 since the initial 6 +-?
– Doug Fir
Sep 17 '18 at 15:49
1
that is not inside the squareroot so we do not include it
– Henry Lee
Sep 17 '18 at 15:51
mathwarehouse.com/dictionary/D-words/… remember where you can apply your commutative and distributive laws
– Henry Lee
Sep 17 '18 at 15:52
add a comment |
so we start off with:
$$x^3+3x^2-4x-4=x^3+2x+2$$
which we can re-write as:
$$3x^2-6x-6=0$$
so we know that:
$$x=frac{-bpmsqrt{b^2-4ac}}{2a}=frac{6pmsqrt{36-(4)(3)(-6)}}{6}=1pmfrac{sqrt{108}}{6}=1pmsqrt{3}$$
The mistake you made was $36-72$ instead of $36+72$
answered Sep 17 '18 at 15:38
Henry Lee
1,713218
Thanks but would 108 not be 114 since the initial 6 +-?
– Doug Fir
Sep 17 '18 at 15:49
1
that is not inside the squareroot so we do not include it
– Henry Lee
Sep 17 '18 at 15:51
mathwarehouse.com/dictionary/D-words/… remember where you can apply your commutative and distributive laws
– Henry Lee
Sep 17 '18 at 15:52
add a comment |
so we start off with:
$$x^3+3x^2-4x-4=x^3+2x+2$$
which we can re-write as:
$$3x^2-6x-6=0$$
so we know that:
$$x=frac{-bpmsqrt{b^2-4ac}}{2a}=frac{6pmsqrt{36-(4)(3)(-6)}}{6}=1pmfrac{sqrt{108}}{6}=1pmsqrt{3}$$
The mistake you made was $36-72$ instead of $36+72$
answered Sep 17 '18 at 15:38
Henry Lee
1,713218
so we start off with:
$$x^3+3x^2-4x-4=x^3+2x+2$$
which we can re-write as:
$$3x^2-6x-6=0$$
so we know that:
$$x=frac{-bpmsqrt{b^2-4ac}}{2a}=frac{6pmsqrt{36-(4)(3)(-6)}}{6}=1pmfrac{sqrt{108}}{6}=1pmsqrt{3}$$
The mistake you made was $36-72$ instead of $36+72$
answered Sep 17 '18 at 15:38
Henry Lee
1,713218
answered Sep 17 '18 at 15:38
Henry Lee
1,713218
answered Sep 17 '18 at 15:38
Henry Lee
1,713218
answered Sep 17 '18 at 15:38
Henry Lee
1,713218
1,713218
Thanks but would 108 not be 114 since the initial 6 +-?
– Doug Fir
Sep 17 '18 at 15:49
1
that is not inside the squareroot so we do not include it
– Henry Lee
Sep 17 '18 at 15:51
mathwarehouse.com/dictionary/D-words/… remember where you can apply your commutative and distributive laws
– Henry Lee
Sep 17 '18 at 15:52
add a comment |
Thanks but would 108 not be 114 since the initial 6 +-?
– Doug Fir
Sep 17 '18 at 15:49
1
that is not inside the squareroot so we do not include it
– Henry Lee
Sep 17 '18 at 15:51
mathwarehouse.com/dictionary/D-words/… remember where you can apply your commutative and distributive laws
– Henry Lee
Sep 17 '18 at 15:52
Thanks but would 108 not be 114 since the initial 6 +-?
– Doug Fir
Sep 17 '18 at 15:49
Thanks but would 108 not be 114 since the initial 6 +-?
– Doug Fir
Sep 17 '18 at 15:49
1
1
that is not inside the squareroot so we do not include it
– Henry Lee
Sep 17 '18 at 15:51
that is not inside the squareroot so we do not include it
– Henry Lee
Sep 17 '18 at 15:51
mathwarehouse.com/dictionary/D-words/… remember where you can apply your commutative and distributive laws
– Henry Lee
Sep 17 '18 at 15:52
mathwarehouse.com/dictionary/D-words/… remember where you can apply your commutative and distributive laws
– Henry Lee
Sep 17 '18 at 15:52
add a comment |
$$x^3+3x^2-4x-4=x^3+2x+2$$
$$3x^2-6x-6=0$$
$$x^2-2x-2=0$$
$$x=frac{-(-2)pmsqrt{(-2)^2-4(1)(-2)}}{2(1)}$$
$$x=frac{2pmsqrt{4+8}}{2}=frac{2pm 2sqrt 3}{2}=1pmsqrt 3$$
answered Sep 17 '18 at 15:44
Deepesh Meena
4,29921025
add a comment |
$$x^3+3x^2-4x-4=x^3+2x+2$$
$$3x^2-6x-6=0$$
$$x^2-2x-2=0$$
$$x=frac{-(-2)pmsqrt{(-2)^2-4(1)(-2)}}{2(1)}$$
$$x=frac{2pmsqrt{4+8}}{2}=frac{2pm 2sqrt 3}{2}=1pmsqrt 3$$
answered Sep 17 '18 at 15:44
Deepesh Meena
4,29921025
add a comment |
$$x^3+3x^2-4x-4=x^3+2x+2$$
$$3x^2-6x-6=0$$
$$x^2-2x-2=0$$
$$x=frac{-(-2)pmsqrt{(-2)^2-4(1)(-2)}}{2(1)}$$
$$x=frac{2pmsqrt{4+8}}{2}=frac{2pm 2sqrt 3}{2}=1pmsqrt 3$$
answered Sep 17 '18 at 15:44
Deepesh Meena
4,29921025
$$x^3+3x^2-4x-4=x^3+2x+2$$
$$3x^2-6x-6=0$$
$$x^2-2x-2=0$$
$$x=frac{-(-2)pmsqrt{(-2)^2-4(1)(-2)}}{2(1)}$$
$$x=frac{2pmsqrt{4+8}}{2}=frac{2pm 2sqrt 3}{2}=1pmsqrt 3$$
answered Sep 17 '18 at 15:44
Deepesh Meena
4,29921025
answered Sep 17 '18 at 15:44
Deepesh Meena
4,29921025
answered Sep 17 '18 at 15:44
Deepesh Meena
4,29921025
answered Sep 17 '18 at 15:44
Deepesh Meena
4,29921025
4,29921025
add a comment |
add a comment |
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1
discriminant will be $36+72$, that's your mistake
– Vasya
Sep 17 '18 at 15:34