Condition for two cubic equations to have a pair of common positive repeated roots
I was asked to find the condition that two cubic equations $a_1x^3 + b_1x^2 + c_1x + d_1 = 0$ and $a_2x^3 + b_2x^2 + c_2x + d_2 = 0$ have a pair of common positive repeated roots. The answer given is that
$$begin{vmatrix}
3a_1&2b_1&c_1\
3a_2&2b_2&c_2\
a_1b_2-a_2b_1&a_1c_2-a_2c_1&a_1d_2-a_2d_1
end{vmatrix} = 0$$
The only hint that I am getting is that we have to use the condition for common root of their derivatives since the roots are repeated. The hint is on the basis of the 3,2,1 coefficients in the determinant. But still I have not been able to go any further with that.
algebra-precalculus determinant quadratics
asked 2 days ago
Shubhraneel PalShubhraneel Pal
35329
add a comment |
I was asked to find the condition that two cubic equations $a_1x^3 + b_1x^2 + c_1x + d_1 = 0$ and $a_2x^3 + b_2x^2 + c_2x + d_2 = 0$ have a pair of common positive repeated roots. The answer given is that
$$begin{vmatrix}
3a_1&2b_1&c_1\
3a_2&2b_2&c_2\
a_1b_2-a_2b_1&a_1c_2-a_2c_1&a_1d_2-a_2d_1
end{vmatrix} = 0$$
The only hint that I am getting is that we have to use the condition for common root of their derivatives since the roots are repeated. The hint is on the basis of the 3,2,1 coefficients in the determinant. But still I have not been able to go any further with that.
algebra-precalculus determinant quadratics
asked 2 days ago
Shubhraneel PalShubhraneel Pal
35329
add a comment |
I was asked to find the condition that two cubic equations $a_1x^3 + b_1x^2 + c_1x + d_1 = 0$ and $a_2x^3 + b_2x^2 + c_2x + d_2 = 0$ have a pair of common positive repeated roots. The answer given is that
$$begin{vmatrix}
3a_1&2b_1&c_1\
3a_2&2b_2&c_2\
a_1b_2-a_2b_1&a_1c_2-a_2c_1&a_1d_2-a_2d_1
end{vmatrix} = 0$$
The only hint that I am getting is that we have to use the condition for common root of their derivatives since the roots are repeated. The hint is on the basis of the 3,2,1 coefficients in the determinant. But still I have not been able to go any further with that.
algebra-precalculus determinant quadratics
asked 2 days ago
Shubhraneel PalShubhraneel Pal
35329
I was asked to find the condition that two cubic equations $a_1x^3 + b_1x^2 + c_1x + d_1 = 0$ and $a_2x^3 + b_2x^2 + c_2x + d_2 = 0$ have a pair of common positive repeated roots. The answer given is that
$$begin{vmatrix}
3a_1&2b_1&c_1\
3a_2&2b_2&c_2\
a_1b_2-a_2b_1&a_1c_2-a_2c_1&a_1d_2-a_2d_1
end{vmatrix} = 0$$
The only hint that I am getting is that we have to use the condition for common root of their derivatives since the roots are repeated. The hint is on the basis of the 3,2,1 coefficients in the determinant. But still I have not been able to go any further with that.
algebra-precalculus determinant quadratics
algebra-precalculus determinant quadratics
asked 2 days ago
Shubhraneel PalShubhraneel Pal
35329
asked 2 days ago
Shubhraneel PalShubhraneel Pal
35329
asked 2 days ago
Shubhraneel PalShubhraneel Pal
35329
asked 2 days ago
Shubhraneel PalShubhraneel Pal
35329
asked 2 days ago
Shubhraneel PalShubhraneel Pal
35329
35329
add a comment |
add a comment |
1 Answer
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If $a_1x^3+b_1x^2+c_1x+d=0$ has one of the roots (say $x_0$) repeating, we can write the equation as $$a_1(x-x_0)^2(x-x_1)=0$$Here $x_1$ is the other root. Now take the derivative of the above expression. You get $$a_12(x-x_0)(x-x_1)+a_1(x-x_0)^2=a_1(x-x_0)(2x-2x_0+x-x_1)$$
Notice that $x_0$ is also root of the derivative. So just write out the four equations saying that $x_0$ is root of the initial equations and root of the derivatives:
$$begin{align}a_1x_0^3+b_1x_0^2+c_1x_0+d_1&=0\a_2x_0^3+b_2x_0^2+c_2x_0+d_2&=0\3a_1x_0^2+2b_1x_0+c_1&=0\3a_2x_0^2+2b_2x_0+c_2&=0end{align}$$
We keep the last two equations, and we eliminate $x_0^3$ from the first two. For that we multiply the first equation by $a_2$, the second by $a_1$ and then subtract the two. You get $$(b_1a_2-b_2a_1)x_0^2+(c_1a_2-c_2a_1)x_0+(d_1a_2-d_2a_1)=0$$
Now all you need to show is that you have a solution for these three quadratic equations only if the coefficients are not linearly independent.
answered 2 days ago
AndreiAndrei
11.4k21026
add a comment |
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1 Answer
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If $a_1x^3+b_1x^2+c_1x+d=0$ has one of the roots (say $x_0$) repeating, we can write the equation as $$a_1(x-x_0)^2(x-x_1)=0$$Here $x_1$ is the other root. Now take the derivative of the above expression. You get $$a_12(x-x_0)(x-x_1)+a_1(x-x_0)^2=a_1(x-x_0)(2x-2x_0+x-x_1)$$
Notice that $x_0$ is also root of the derivative. So just write out the four equations saying that $x_0$ is root of the initial equations and root of the derivatives:
$$begin{align}a_1x_0^3+b_1x_0^2+c_1x_0+d_1&=0\a_2x_0^3+b_2x_0^2+c_2x_0+d_2&=0\3a_1x_0^2+2b_1x_0+c_1&=0\3a_2x_0^2+2b_2x_0+c_2&=0end{align}$$
We keep the last two equations, and we eliminate $x_0^3$ from the first two. For that we multiply the first equation by $a_2$, the second by $a_1$ and then subtract the two. You get $$(b_1a_2-b_2a_1)x_0^2+(c_1a_2-c_2a_1)x_0+(d_1a_2-d_2a_1)=0$$
Now all you need to show is that you have a solution for these three quadratic equations only if the coefficients are not linearly independent.
answered 2 days ago
AndreiAndrei
11.4k21026
add a comment |
If $a_1x^3+b_1x^2+c_1x+d=0$ has one of the roots (say $x_0$) repeating, we can write the equation as $$a_1(x-x_0)^2(x-x_1)=0$$Here $x_1$ is the other root. Now take the derivative of the above expression. You get $$a_12(x-x_0)(x-x_1)+a_1(x-x_0)^2=a_1(x-x_0)(2x-2x_0+x-x_1)$$
Notice that $x_0$ is also root of the derivative. So just write out the four equations saying that $x_0$ is root of the initial equations and root of the derivatives:
$$begin{align}a_1x_0^3+b_1x_0^2+c_1x_0+d_1&=0\a_2x_0^3+b_2x_0^2+c_2x_0+d_2&=0\3a_1x_0^2+2b_1x_0+c_1&=0\3a_2x_0^2+2b_2x_0+c_2&=0end{align}$$
We keep the last two equations, and we eliminate $x_0^3$ from the first two. For that we multiply the first equation by $a_2$, the second by $a_1$ and then subtract the two. You get $$(b_1a_2-b_2a_1)x_0^2+(c_1a_2-c_2a_1)x_0+(d_1a_2-d_2a_1)=0$$
Now all you need to show is that you have a solution for these three quadratic equations only if the coefficients are not linearly independent.
answered 2 days ago
AndreiAndrei
11.4k21026
add a comment |
If $a_1x^3+b_1x^2+c_1x+d=0$ has one of the roots (say $x_0$) repeating, we can write the equation as $$a_1(x-x_0)^2(x-x_1)=0$$Here $x_1$ is the other root. Now take the derivative of the above expression. You get $$a_12(x-x_0)(x-x_1)+a_1(x-x_0)^2=a_1(x-x_0)(2x-2x_0+x-x_1)$$
Notice that $x_0$ is also root of the derivative. So just write out the four equations saying that $x_0$ is root of the initial equations and root of the derivatives:
$$begin{align}a_1x_0^3+b_1x_0^2+c_1x_0+d_1&=0\a_2x_0^3+b_2x_0^2+c_2x_0+d_2&=0\3a_1x_0^2+2b_1x_0+c_1&=0\3a_2x_0^2+2b_2x_0+c_2&=0end{align}$$
We keep the last two equations, and we eliminate $x_0^3$ from the first two. For that we multiply the first equation by $a_2$, the second by $a_1$ and then subtract the two. You get $$(b_1a_2-b_2a_1)x_0^2+(c_1a_2-c_2a_1)x_0+(d_1a_2-d_2a_1)=0$$
Now all you need to show is that you have a solution for these three quadratic equations only if the coefficients are not linearly independent.
answered 2 days ago
AndreiAndrei
11.4k21026
If $a_1x^3+b_1x^2+c_1x+d=0$ has one of the roots (say $x_0$) repeating, we can write the equation as $$a_1(x-x_0)^2(x-x_1)=0$$Here $x_1$ is the other root. Now take the derivative of the above expression. You get $$a_12(x-x_0)(x-x_1)+a_1(x-x_0)^2=a_1(x-x_0)(2x-2x_0+x-x_1)$$
Notice that $x_0$ is also root of the derivative. So just write out the four equations saying that $x_0$ is root of the initial equations and root of the derivatives:
$$begin{align}a_1x_0^3+b_1x_0^2+c_1x_0+d_1&=0\a_2x_0^3+b_2x_0^2+c_2x_0+d_2&=0\3a_1x_0^2+2b_1x_0+c_1&=0\3a_2x_0^2+2b_2x_0+c_2&=0end{align}$$
We keep the last two equations, and we eliminate $x_0^3$ from the first two. For that we multiply the first equation by $a_2$, the second by $a_1$ and then subtract the two. You get $$(b_1a_2-b_2a_1)x_0^2+(c_1a_2-c_2a_1)x_0+(d_1a_2-d_2a_1)=0$$
Now all you need to show is that you have a solution for these three quadratic equations only if the coefficients are not linearly independent.
answered 2 days ago
AndreiAndrei
11.4k21026
answered 2 days ago
AndreiAndrei
11.4k21026
answered 2 days ago
AndreiAndrei
11.4k21026
answered 2 days ago
AndreiAndrei
11.4k21026
11.4k21026
add a comment |
add a comment |
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