Question

x^4-10x^3+26x^2-10x+1=0 solve this equation by ferrari method​

Answer 1

Solve this $x^4-10x^3+26x^2-10x+1=0$   with help of the blow method (ferrari method)

Step-by-step explanation:

Let that polynomial equation:

P (x). = $x^4-10x^3+26x^2-10x+1=0$

Put x = 1

P (1) = 1–10+35–50+24=60–60=0, so, x=1 must be a root — because P (1)=0.

P (x) = $x^{4} - x^{3} - 9 x^{3} + 9 x^{2} + 26 x^{2} - 26x -24x+24 = 0$

$P (x)= x^{3} (x-1)-9 x^{2} (x-1)+26x(x-1)-24(x-1) = 0$

$P (x)=(x-1)( x^{3} - 9 x^{2}+ 26x-24)$

Look for easy roots for that 3rd degree polynomial equation:

$P(x) = ( x^{3} -9 x^{2} + 26x-24)$

P(2) = 8–36+52–24=0, Therefore P(2) = 0.

Factorise  (x-2) out of it:

$P(x) = (x-1) (x^ {3}-2 x^ {2}- 7 x^ {2} +14x+12x- 24)$

$P(x) = (x-1) (x-2) (x^ {2}-7x+12)$

$P(x) = (x-1)(x-2) ( x^{2}-3x - 4x +12)$

$P(x) = (x-1)(x-2) [( x(x-3) - 4(x - 3)]$

$P(x) = (x-1)(x-2) (x - 4 )(x-3)$

You end up with 4 solutions for the equation P(x)=0, and they are: 1, 2, 3, and 4.

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Answer 2

To Solve : For values of x

(By Lodovico Ferrari's Approach) such that:

$\\ f(x) = {x}^{4} - 10 {x}^{3} + 26 {x}^{2} - 10x + 1 = 0$

Completing the square once:

$\\ ( {x}^{2} ) ^{2} - 2(5x)( {x}^{2} ) + (5x)^{2} + {x}^{2} - 10x + 1 = 0$

$( {x}^{2} - 5 x) ^{2} + {x}^{2} - 10x + 1 = 0$

Completing the Square Twice by introducing a new variable:

$\\ ( {x}^{2} - 5x) ^{2} + 2z( {x}^{2} - 5x) + {z}^{2} + (1 - 2z) {x}^{2} + 10(z - 1)x + (1 - {z}^{2} ) = 0$

$\\ ( {x}^{2} - 5x + z) ^{2} - ( (2z - 1){x}^{2} + 10x(1 - z) + ( {z}^{2} - 1)) = 0$

Taking Rest of the terms To the right hand side:

$\\ ( {x}^{2} - 5x + z) ^{2} = (2z - 1) {x}^{2} + 10x(1 - z) + ( {z}^{2} - 1)$

Now, For chosing the real value of the new variable such that the right hand side also becomes a perfect square.

Quadratic Discriminant of the Polynomial on RHS must be zero.

Therefore,

$\\ 4( {z}^{2} - 1)(2z - 1) - {10}^{2} {(1 - z)}^{2} = 0$

$8 {z}^{3} - 104 {z}^{2} + 192z - 96 = 0$

${z}^{3} - 13 {z}^{2} + 24z - 12 = 0$

By Rational Zeroes Theorem , we get one of the simplest rational root as : z = + 1

Therefore, the previous equation becomes:

${( {x}^{2} - 5x + 1) }= \pm (x)$

$\\ \therefore f(x) = ( {x}^{2} - 6x + 1)( {x}^{2} - 4x + 1) = 0$

Using the Quadratic formula, we can solve for all the values x, that are:

$\implies \: x_1 = 2 - \sqrt{3}$

$\implies \:x_2 = 2 + \sqrt{ 3}$

$\implies \:x_3 = 3 - 2 \sqrt{2}$

$\implies \:x_4 = 3 + 2 \sqrt{2}$

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