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If the roots of the equation $x^2 + 3x -1 = 0$ also satisfy the equation $x^4 + ax^2 + bx + c = 0$ find the value of $a + b + 4c + 100$
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If the roots of the equation $x^2 + 3x -1 = 0$ also satisfy the equation $x^4 + ax^2 + bx + c = 0$ find the value of $a + b + 4c + 100$
I tried really hard but the most I could get is the sum of the roots of the second equation is $3$
Please could someone solve this please !
It would mean the world to me
algebra-precalculus polynomials roots
edited Aug 6 at 14:26
TheSimpliFire
9,69261952
asked Aug 6 at 14:09
John Tom
816
add a comment |Â
up vote
1
down vote
favorite
If the roots of the equation $x^2 + 3x -1 = 0$ also satisfy the equation $x^4 + ax^2 + bx + c = 0$ find the value of $a + b + 4c + 100$
I tried really hard but the most I could get is the sum of the roots of the second equation is $3$
Please could someone solve this please !
It would mean the world to me
algebra-precalculus polynomials roots
edited Aug 6 at 14:26
TheSimpliFire
9,69261952
asked Aug 6 at 14:09
John Tom
816
The discriminant of $x^2+3x-1$ is non-zero. Therefore its two roots are different. This means that if they satisfy $x^4+ax^2+bx+c$ it must be because this polynomial is divisible by $x^2+3x-1$. Dividing you get remainder $(-3a +b-33)x + (a + c + 10)$. Therefore both coefficients should be $0$, i.e. $-3a +b-33=0$ and $a + c + 10=0$. Add the first equation to $4$ times the second to get $a+b+4c+7=0$. Adding $93$ on both sides $a+b+4c+100=93$.
– user580373
Aug 6 at 14:28
I didn't get the coefficient being 0 part .could you explain that part?
– John Tom
Aug 6 at 15:56
If a polynomial divides another, then remainder of the division should be the zero polynomial.
– user580373
Aug 6 at 15:59
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If the roots of the equation $x^2 + 3x -1 = 0$ also satisfy the equation $x^4 + ax^2 + bx + c = 0$ find the value of $a + b + 4c + 100$
I tried really hard but the most I could get is the sum of the roots of the second equation is $3$
Please could someone solve this please !
It would mean the world to me
algebra-precalculus polynomials roots
edited Aug 6 at 14:26
TheSimpliFire
9,69261952
asked Aug 6 at 14:09
John Tom
816
If the roots of the equation $x^2 + 3x -1 = 0$ also satisfy the equation $x^4 + ax^2 + bx + c = 0$ find the value of $a + b + 4c + 100$
I tried really hard but the most I could get is the sum of the roots of the second equation is $3$
Please could someone solve this please !
It would mean the world to me
algebra-precalculus polynomials roots
edited Aug 6 at 14:26
TheSimpliFire
9,69261952
asked Aug 6 at 14:09
John Tom
816
edited Aug 6 at 14:26
TheSimpliFire
9,69261952
edited Aug 6 at 14:26
TheSimpliFire
9,69261952
edited Aug 6 at 14:26
TheSimpliFire
9,69261952
9,69261952
asked Aug 6 at 14:09
John Tom
816
asked Aug 6 at 14:09
John Tom
816
asked Aug 6 at 14:09
John Tom
816
816
The discriminant of $x^2+3x-1$ is non-zero. Therefore its two roots are different. This means that if they satisfy $x^4+ax^2+bx+c$ it must be because this polynomial is divisible by $x^2+3x-1$. Dividing you get remainder $(-3a +b-33)x + (a + c + 10)$. Therefore both coefficients should be $0$, i.e. $-3a +b-33=0$ and $a + c + 10=0$. Add the first equation to $4$ times the second to get $a+b+4c+7=0$. Adding $93$ on both sides $a+b+4c+100=93$.
– user580373
Aug 6 at 14:28
I didn't get the coefficient being 0 part .could you explain that part?
– John Tom
Aug 6 at 15:56
If a polynomial divides another, then remainder of the division should be the zero polynomial.
– user580373
Aug 6 at 15:59
add a comment |Â
The discriminant of $x^2+3x-1$ is non-zero. Therefore its two roots are different. This means that if they satisfy $x^4+ax^2+bx+c$ it must be because this polynomial is divisible by $x^2+3x-1$. Dividing you get remainder $(-3a +b-33)x + (a + c + 10)$. Therefore both coefficients should be $0$, i.e. $-3a +b-33=0$ and $a + c + 10=0$. Add the first equation to $4$ times the second to get $a+b+4c+7=0$. Adding $93$ on both sides $a+b+4c+100=93$.
– user580373
Aug 6 at 14:28
I didn't get the coefficient being 0 part .could you explain that part?
– John Tom
Aug 6 at 15:56
If a polynomial divides another, then remainder of the division should be the zero polynomial.
– user580373
Aug 6 at 15:59
The discriminant of $x^2+3x-1$ is non-zero. Therefore its two roots are different. This means that if they satisfy $x^4+ax^2+bx+c$ it must be because this polynomial is divisible by $x^2+3x-1$. Dividing you get remainder $(-3a +b-33)x + (a + c + 10)$. Therefore both coefficients should be $0$, i.e. $-3a +b-33=0$ and $a + c + 10=0$. Add the first equation to $4$ times the second to get $a+b+4c+7=0$. Adding $93$ on both sides $a+b+4c+100=93$.
– user580373
Aug 6 at 14:28
The discriminant of $x^2+3x-1$ is non-zero. Therefore its two roots are different. This means that if they satisfy $x^4+ax^2+bx+c$ it must be because this polynomial is divisible by $x^2+3x-1$. Dividing you get remainder $(-3a +b-33)x + (a + c + 10)$. Therefore both coefficients should be $0$, i.e. $-3a +b-33=0$ and $a + c + 10=0$. Add the first equation to $4$ times the second to get $a+b+4c+7=0$. Adding $93$ on both sides $a+b+4c+100=93$.
– user580373
Aug 6 at 14:28
I didn't get the coefficient being 0 part .could you explain that part?
– John Tom
Aug 6 at 15:56
I didn't get the coefficient being 0 part .could you explain that part?
– John Tom
Aug 6 at 15:56
If a polynomial divides another, then remainder of the division should be the zero polynomial.
– user580373
Aug 6 at 15:59
If a polynomial divides another, then remainder of the division should be the zero polynomial.
– user580373
Aug 6 at 15:59
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
2
down vote
We can find the equation must satisfy$x^4+ax^2+bx+c=(x^2+3x-1)(x^2+mx+n)$ and$m=-3,n=-c$. So when$x=1,1+a+b+c=3(1-3-c)rightarrow a+b+4c+7=0$.
answered Aug 6 at 14:22
HQR
298111
thank you @HQR @ Dr.Sonnhard Graubner @ TheSimpliFire THANK YOU
– John Tom
Aug 6 at 15:10
add a comment |Â
up vote
1
down vote
Hint: The solutions of your polynomial are given by
$$x_1,2=-frac32pmfracsqrt132$$
plugging $x_1$ into the other equation we get
$$bsqrt13+3csqrt13-3sqrt13+2a+3b+11c+11=0$$
the rest is for you!
answered Aug 6 at 14:21
Dr. Sonnhard Graubner
67k32660
Does this approach lead to the solution? What should be the next step?
– Jaroslaw Matlak
Aug 6 at 15:25
This have i written already
– Dr. Sonnhard Graubner
Aug 6 at 15:29
1
You must plug in the other solution, $x_2$
– Dr. Sonnhard Graubner
Aug 6 at 15:30
...and i'll obtain the system of 2 linear equations, from which I can get $a(c)$ and $b(c)$? Now it's clear, thank you :)
– Jaroslaw Matlak
Aug 6 at 15:55
This is nice, that i could help you!
– Dr. Sonnhard Graubner
Aug 6 at 15:56
add a comment |Â
up vote
1
down vote
Alternatively, the Vieta's formula's for $x^2 + 3x -1 = 0$:
$$x_1+x_2=-3, x_1x_2=-1 Rightarrow x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=9+2=11.$$
Plugging $x_1,x_2$ to $x^4 + ax^2 + bx + c = 0$ and subtracting:
$$begincasesx_1^4 + ax_1^2 + bx_1 + c = 0\ x_2^4 + ax_2^2 + bx_2 + c = 0endcases Rightarrow \
(x_1-x_2)(x_1+x_2)(x_1^2+x_2^2)+a(x_1-x_2)(x_1+x_2)+b(x_1-x_2)=0 Rightarrow \
requirecancelcancel(x_1-x_2)[(-3)(11)+a(-3)+b]=0 Rightarrow \ 3a-b=-33 (1)$$
and adding:
$$(x_1^2+x_2^2)^2-2(x_1x_2)^2+a(x_1^2+x_2^2)+b(x_1+x_2)+2c=0 Rightarrow \
11^2-2+11a-3b+2c=0 Rightarrow \
119+2a+3(underbrace3a-b_=-33)+2c=0 Rightarrow \
a+c=-10 (2)$$
Multiply $(2)$ by $4$ and subtract $(1)$ to get:
$$a+b+4c=-7 Rightarrow a+b+4c+100=93.$$
answered Aug 6 at 15:12
farruhota
13.8k2632
add a comment |Â
up vote
0
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We have:
$$(x^2+3x-1)(x^2+px+q)=x^4+ax^2+bx+c$$
We can expand the left side:
$$x^4+x^3(3+p)+x^2(q+3p-1)+x(3q-p)-q=x^4+ax^2+bx+c$$
Because there is no $x^3$ element on the right side, we conclude
$$3+p=0$$
$$p=-3$$
We have then:
$$x^4+x^2(q-10)+x(3q+3)-q=x^4+ax^2+bx+c$$
thus:
$$begincasesa=q-10 \ b=3q+3 \ c=-qendcases$$
Subtracting these into $a+b+4c+100$ we obtain:
$$a+b+4c+100=q-10+3q+3-4q+100 = -7+100=93$$
answered Aug 6 at 15:22
Jaroslaw Matlak
3,880830
thank you Jaroslaw Matlak for the complete explanation
– John Tom
Aug 6 at 15:59
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
We can find the equation must satisfy$x^4+ax^2+bx+c=(x^2+3x-1)(x^2+mx+n)$ and$m=-3,n=-c$. So when$x=1,1+a+b+c=3(1-3-c)rightarrow a+b+4c+7=0$.
answered Aug 6 at 14:22
HQR
298111
thank you @HQR @ Dr.Sonnhard Graubner @ TheSimpliFire THANK YOU
– John Tom
Aug 6 at 15:10
add a comment |Â
up vote
2
down vote
We can find the equation must satisfy$x^4+ax^2+bx+c=(x^2+3x-1)(x^2+mx+n)$ and$m=-3,n=-c$. So when$x=1,1+a+b+c=3(1-3-c)rightarrow a+b+4c+7=0$.
answered Aug 6 at 14:22
HQR
298111
thank you @HQR @ Dr.Sonnhard Graubner @ TheSimpliFire THANK YOU
– John Tom
Aug 6 at 15:10
add a comment |Â
up vote
2
down vote
up vote
2
down vote
We can find the equation must satisfy$x^4+ax^2+bx+c=(x^2+3x-1)(x^2+mx+n)$ and$m=-3,n=-c$. So when$x=1,1+a+b+c=3(1-3-c)rightarrow a+b+4c+7=0$.
answered Aug 6 at 14:22
HQR
298111
We can find the equation must satisfy$x^4+ax^2+bx+c=(x^2+3x-1)(x^2+mx+n)$ and$m=-3,n=-c$. So when$x=1,1+a+b+c=3(1-3-c)rightarrow a+b+4c+7=0$.
answered Aug 6 at 14:22
HQR
298111
answered Aug 6 at 14:22
HQR
298111
answered Aug 6 at 14:22
HQR
298111
answered Aug 6 at 14:22
HQR
298111
298111
thank you @HQR @ Dr.Sonnhard Graubner @ TheSimpliFire THANK YOU
– John Tom
Aug 6 at 15:10
add a comment |Â
thank you @HQR @ Dr.Sonnhard Graubner @ TheSimpliFire THANK YOU
– John Tom
Aug 6 at 15:10
thank you @HQR @ Dr.Sonnhard Graubner @ TheSimpliFire THANK YOU
– John Tom
Aug 6 at 15:10
thank you @HQR @ Dr.Sonnhard Graubner @ TheSimpliFire THANK YOU
– John Tom
Aug 6 at 15:10
add a comment |Â
up vote
1
down vote
Hint: The solutions of your polynomial are given by
$$x_1,2=-frac32pmfracsqrt132$$
plugging $x_1$ into the other equation we get
$$bsqrt13+3csqrt13-3sqrt13+2a+3b+11c+11=0$$
the rest is for you!
answered Aug 6 at 14:21
Dr. Sonnhard Graubner
67k32660
Does this approach lead to the solution? What should be the next step?
– Jaroslaw Matlak
Aug 6 at 15:25
This have i written already
– Dr. Sonnhard Graubner
Aug 6 at 15:29
1
You must plug in the other solution, $x_2$
– Dr. Sonnhard Graubner
Aug 6 at 15:30
...and i'll obtain the system of 2 linear equations, from which I can get $a(c)$ and $b(c)$? Now it's clear, thank you :)
– Jaroslaw Matlak
Aug 6 at 15:55
This is nice, that i could help you!
– Dr. Sonnhard Graubner
Aug 6 at 15:56
add a comment |Â
up vote
1
down vote
Hint: The solutions of your polynomial are given by
$$x_1,2=-frac32pmfracsqrt132$$
plugging $x_1$ into the other equation we get
$$bsqrt13+3csqrt13-3sqrt13+2a+3b+11c+11=0$$
the rest is for you!
answered Aug 6 at 14:21
Dr. Sonnhard Graubner
67k32660
Does this approach lead to the solution? What should be the next step?
– Jaroslaw Matlak
Aug 6 at 15:25
This have i written already
– Dr. Sonnhard Graubner
Aug 6 at 15:29
1
You must plug in the other solution, $x_2$
– Dr. Sonnhard Graubner
Aug 6 at 15:30
...and i'll obtain the system of 2 linear equations, from which I can get $a(c)$ and $b(c)$? Now it's clear, thank you :)
– Jaroslaw Matlak
Aug 6 at 15:55
This is nice, that i could help you!
– Dr. Sonnhard Graubner
Aug 6 at 15:56
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint: The solutions of your polynomial are given by
$$x_1,2=-frac32pmfracsqrt132$$
plugging $x_1$ into the other equation we get
$$bsqrt13+3csqrt13-3sqrt13+2a+3b+11c+11=0$$
the rest is for you!
answered Aug 6 at 14:21
Dr. Sonnhard Graubner
67k32660
Hint: The solutions of your polynomial are given by
$$x_1,2=-frac32pmfracsqrt132$$
plugging $x_1$ into the other equation we get
$$bsqrt13+3csqrt13-3sqrt13+2a+3b+11c+11=0$$
the rest is for you!
answered Aug 6 at 14:21
Dr. Sonnhard Graubner
67k32660
answered Aug 6 at 14:21
Dr. Sonnhard Graubner
67k32660
answered Aug 6 at 14:21
Dr. Sonnhard Graubner
67k32660
answered Aug 6 at 14:21
Dr. Sonnhard Graubner
67k32660
67k32660
Does this approach lead to the solution? What should be the next step?
– Jaroslaw Matlak
Aug 6 at 15:25
This have i written already
– Dr. Sonnhard Graubner
Aug 6 at 15:29
1
You must plug in the other solution, $x_2$
– Dr. Sonnhard Graubner
Aug 6 at 15:30
...and i'll obtain the system of 2 linear equations, from which I can get $a(c)$ and $b(c)$? Now it's clear, thank you :)
– Jaroslaw Matlak
Aug 6 at 15:55
This is nice, that i could help you!
– Dr. Sonnhard Graubner
Aug 6 at 15:56
add a comment |Â
Does this approach lead to the solution? What should be the next step?
– Jaroslaw Matlak
Aug 6 at 15:25
This have i written already
– Dr. Sonnhard Graubner
Aug 6 at 15:29
1
You must plug in the other solution, $x_2$
– Dr. Sonnhard Graubner
Aug 6 at 15:30
...and i'll obtain the system of 2 linear equations, from which I can get $a(c)$ and $b(c)$? Now it's clear, thank you :)
– Jaroslaw Matlak
Aug 6 at 15:55
This is nice, that i could help you!
– Dr. Sonnhard Graubner
Aug 6 at 15:56
Does this approach lead to the solution? What should be the next step?
– Jaroslaw Matlak
Aug 6 at 15:25
Does this approach lead to the solution? What should be the next step?
– Jaroslaw Matlak
Aug 6 at 15:25
This have i written already
– Dr. Sonnhard Graubner
Aug 6 at 15:29
This have i written already
– Dr. Sonnhard Graubner
Aug 6 at 15:29
1
1
You must plug in the other solution, $x_2$
– Dr. Sonnhard Graubner
Aug 6 at 15:30
You must plug in the other solution, $x_2$
– Dr. Sonnhard Graubner
Aug 6 at 15:30
...and i'll obtain the system of 2 linear equations, from which I can get $a(c)$ and $b(c)$? Now it's clear, thank you :)
– Jaroslaw Matlak
Aug 6 at 15:55
...and i'll obtain the system of 2 linear equations, from which I can get $a(c)$ and $b(c)$? Now it's clear, thank you :)
– Jaroslaw Matlak
Aug 6 at 15:55
This is nice, that i could help you!
– Dr. Sonnhard Graubner
Aug 6 at 15:56
This is nice, that i could help you!
– Dr. Sonnhard Graubner
Aug 6 at 15:56
add a comment |Â
up vote
1
down vote
Alternatively, the Vieta's formula's for $x^2 + 3x -1 = 0$:
$$x_1+x_2=-3, x_1x_2=-1 Rightarrow x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=9+2=11.$$
Plugging $x_1,x_2$ to $x^4 + ax^2 + bx + c = 0$ and subtracting:
$$begincasesx_1^4 + ax_1^2 + bx_1 + c = 0\ x_2^4 + ax_2^2 + bx_2 + c = 0endcases Rightarrow \
(x_1-x_2)(x_1+x_2)(x_1^2+x_2^2)+a(x_1-x_2)(x_1+x_2)+b(x_1-x_2)=0 Rightarrow \
requirecancelcancel(x_1-x_2)[(-3)(11)+a(-3)+b]=0 Rightarrow \ 3a-b=-33 (1)$$
and adding:
$$(x_1^2+x_2^2)^2-2(x_1x_2)^2+a(x_1^2+x_2^2)+b(x_1+x_2)+2c=0 Rightarrow \
11^2-2+11a-3b+2c=0 Rightarrow \
119+2a+3(underbrace3a-b_=-33)+2c=0 Rightarrow \
a+c=-10 (2)$$
Multiply $(2)$ by $4$ and subtract $(1)$ to get:
$$a+b+4c=-7 Rightarrow a+b+4c+100=93.$$
answered Aug 6 at 15:12
farruhota
13.8k2632
add a comment |Â
up vote
1
down vote
Alternatively, the Vieta's formula's for $x^2 + 3x -1 = 0$:
$$x_1+x_2=-3, x_1x_2=-1 Rightarrow x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=9+2=11.$$
Plugging $x_1,x_2$ to $x^4 + ax^2 + bx + c = 0$ and subtracting:
$$begincasesx_1^4 + ax_1^2 + bx_1 + c = 0\ x_2^4 + ax_2^2 + bx_2 + c = 0endcases Rightarrow \
(x_1-x_2)(x_1+x_2)(x_1^2+x_2^2)+a(x_1-x_2)(x_1+x_2)+b(x_1-x_2)=0 Rightarrow \
requirecancelcancel(x_1-x_2)[(-3)(11)+a(-3)+b]=0 Rightarrow \ 3a-b=-33 (1)$$
and adding:
$$(x_1^2+x_2^2)^2-2(x_1x_2)^2+a(x_1^2+x_2^2)+b(x_1+x_2)+2c=0 Rightarrow \
11^2-2+11a-3b+2c=0 Rightarrow \
119+2a+3(underbrace3a-b_=-33)+2c=0 Rightarrow \
a+c=-10 (2)$$
Multiply $(2)$ by $4$ and subtract $(1)$ to get:
$$a+b+4c=-7 Rightarrow a+b+4c+100=93.$$
answered Aug 6 at 15:12
farruhota
13.8k2632
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Alternatively, the Vieta's formula's for $x^2 + 3x -1 = 0$:
$$x_1+x_2=-3, x_1x_2=-1 Rightarrow x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=9+2=11.$$
Plugging $x_1,x_2$ to $x^4 + ax^2 + bx + c = 0$ and subtracting:
$$begincasesx_1^4 + ax_1^2 + bx_1 + c = 0\ x_2^4 + ax_2^2 + bx_2 + c = 0endcases Rightarrow \
(x_1-x_2)(x_1+x_2)(x_1^2+x_2^2)+a(x_1-x_2)(x_1+x_2)+b(x_1-x_2)=0 Rightarrow \
requirecancelcancel(x_1-x_2)[(-3)(11)+a(-3)+b]=0 Rightarrow \ 3a-b=-33 (1)$$
and adding:
$$(x_1^2+x_2^2)^2-2(x_1x_2)^2+a(x_1^2+x_2^2)+b(x_1+x_2)+2c=0 Rightarrow \
11^2-2+11a-3b+2c=0 Rightarrow \
119+2a+3(underbrace3a-b_=-33)+2c=0 Rightarrow \
a+c=-10 (2)$$
Multiply $(2)$ by $4$ and subtract $(1)$ to get:
$$a+b+4c=-7 Rightarrow a+b+4c+100=93.$$
answered Aug 6 at 15:12
farruhota
13.8k2632
Alternatively, the Vieta's formula's for $x^2 + 3x -1 = 0$:
$$x_1+x_2=-3, x_1x_2=-1 Rightarrow x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=9+2=11.$$
Plugging $x_1,x_2$ to $x^4 + ax^2 + bx + c = 0$ and subtracting:
$$begincasesx_1^4 + ax_1^2 + bx_1 + c = 0\ x_2^4 + ax_2^2 + bx_2 + c = 0endcases Rightarrow \
(x_1-x_2)(x_1+x_2)(x_1^2+x_2^2)+a(x_1-x_2)(x_1+x_2)+b(x_1-x_2)=0 Rightarrow \
requirecancelcancel(x_1-x_2)[(-3)(11)+a(-3)+b]=0 Rightarrow \ 3a-b=-33 (1)$$
and adding:
$$(x_1^2+x_2^2)^2-2(x_1x_2)^2+a(x_1^2+x_2^2)+b(x_1+x_2)+2c=0 Rightarrow \
11^2-2+11a-3b+2c=0 Rightarrow \
119+2a+3(underbrace3a-b_=-33)+2c=0 Rightarrow \
a+c=-10 (2)$$
Multiply $(2)$ by $4$ and subtract $(1)$ to get:
$$a+b+4c=-7 Rightarrow a+b+4c+100=93.$$
answered Aug 6 at 15:12
farruhota
13.8k2632
answered Aug 6 at 15:12
farruhota
13.8k2632
answered Aug 6 at 15:12
farruhota
13.8k2632
answered Aug 6 at 15:12
farruhota
13.8k2632
13.8k2632
add a comment |Â
add a comment |Â
up vote
0
down vote
We have:
$$(x^2+3x-1)(x^2+px+q)=x^4+ax^2+bx+c$$
We can expand the left side:
$$x^4+x^3(3+p)+x^2(q+3p-1)+x(3q-p)-q=x^4+ax^2+bx+c$$
Because there is no $x^3$ element on the right side, we conclude
$$3+p=0$$
$$p=-3$$
We have then:
$$x^4+x^2(q-10)+x(3q+3)-q=x^4+ax^2+bx+c$$
thus:
$$begincasesa=q-10 \ b=3q+3 \ c=-qendcases$$
Subtracting these into $a+b+4c+100$ we obtain:
$$a+b+4c+100=q-10+3q+3-4q+100 = -7+100=93$$
answered Aug 6 at 15:22
Jaroslaw Matlak
3,880830
thank you Jaroslaw Matlak for the complete explanation
– John Tom
Aug 6 at 15:59
add a comment |Â
up vote
0
down vote
We have:
$$(x^2+3x-1)(x^2+px+q)=x^4+ax^2+bx+c$$
We can expand the left side:
$$x^4+x^3(3+p)+x^2(q+3p-1)+x(3q-p)-q=x^4+ax^2+bx+c$$
Because there is no $x^3$ element on the right side, we conclude
$$3+p=0$$
$$p=-3$$
We have then:
$$x^4+x^2(q-10)+x(3q+3)-q=x^4+ax^2+bx+c$$
thus:
$$begincasesa=q-10 \ b=3q+3 \ c=-qendcases$$
Subtracting these into $a+b+4c+100$ we obtain:
$$a+b+4c+100=q-10+3q+3-4q+100 = -7+100=93$$
answered Aug 6 at 15:22
Jaroslaw Matlak
3,880830
thank you Jaroslaw Matlak for the complete explanation
– John Tom
Aug 6 at 15:59
add a comment |Â
up vote
0
down vote
up vote
0
down vote
We have:
$$(x^2+3x-1)(x^2+px+q)=x^4+ax^2+bx+c$$
We can expand the left side:
$$x^4+x^3(3+p)+x^2(q+3p-1)+x(3q-p)-q=x^4+ax^2+bx+c$$
Because there is no $x^3$ element on the right side, we conclude
$$3+p=0$$
$$p=-3$$
We have then:
$$x^4+x^2(q-10)+x(3q+3)-q=x^4+ax^2+bx+c$$
thus:
$$begincasesa=q-10 \ b=3q+3 \ c=-qendcases$$
Subtracting these into $a+b+4c+100$ we obtain:
$$a+b+4c+100=q-10+3q+3-4q+100 = -7+100=93$$
answered Aug 6 at 15:22
Jaroslaw Matlak
3,880830
We have:
$$(x^2+3x-1)(x^2+px+q)=x^4+ax^2+bx+c$$
We can expand the left side:
$$x^4+x^3(3+p)+x^2(q+3p-1)+x(3q-p)-q=x^4+ax^2+bx+c$$
Because there is no $x^3$ element on the right side, we conclude
$$3+p=0$$
$$p=-3$$
We have then:
$$x^4+x^2(q-10)+x(3q+3)-q=x^4+ax^2+bx+c$$
thus:
$$begincasesa=q-10 \ b=3q+3 \ c=-qendcases$$
Subtracting these into $a+b+4c+100$ we obtain:
$$a+b+4c+100=q-10+3q+3-4q+100 = -7+100=93$$
answered Aug 6 at 15:22
Jaroslaw Matlak
3,880830
answered Aug 6 at 15:22
Jaroslaw Matlak
3,880830
answered Aug 6 at 15:22
Jaroslaw Matlak
3,880830
answered Aug 6 at 15:22
Jaroslaw Matlak
3,880830
3,880830
thank you Jaroslaw Matlak for the complete explanation
– John Tom
Aug 6 at 15:59
add a comment |Â
thank you Jaroslaw Matlak for the complete explanation
– John Tom
Aug 6 at 15:59
thank you Jaroslaw Matlak for the complete explanation
– John Tom
Aug 6 at 15:59
thank you Jaroslaw Matlak for the complete explanation
– John Tom
Aug 6 at 15:59
add a comment |Â
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The discriminant of $x^2+3x-1$ is non-zero. Therefore its two roots are different. This means that if they satisfy $x^4+ax^2+bx+c$ it must be because this polynomial is divisible by $x^2+3x-1$. Dividing you get remainder $(-3a +b-33)x + (a + c + 10)$. Therefore both coefficients should be $0$, i.e. $-3a +b-33=0$ and $a + c + 10=0$. Add the first equation to $4$ times the second to get $a+b+4c+7=0$. Adding $93$ on both sides $a+b+4c+100=93$.
– user580373
Aug 6 at 14:28
I didn't get the coefficient being 0 part .could you explain that part?
– John Tom
Aug 6 at 15:56
If a polynomial divides another, then remainder of the division should be the zero polynomial.
– user580373
Aug 6 at 15:59