Question

x^4+x^3-4x^2+x+1=0 solve by using qyadratic eqyations​

Answer 1

$\large\underline{\sf{Solution-}}$

Given equation is

$\rm :\longmapsto\: {x}^{4} + {x}^{3} - {4x}^{2} + x + 1 = 0$

can be rewritten as

$\rm :\longmapsto\: {x}^{4} + {x}^{3} \red{ - {2x}^{2} - {2x}^{2}} + x + 1 = 0$

can be re-arranged as

$\rm :\longmapsto\: {x}^{4} - {2x}^{2} + 1 + {x}^{3} - {2x}^{2} + x = 0$

$\rm :\longmapsto\: [{x}^{4} - {2x}^{2} + 1 ]+ x[{x}^{2} - {2x}^{} + 1] = 0$

$\rm :\longmapsto\: [{ {(x}^{2} )}^{2} - {2x}^{2} + {1}^{2} ]+ x[{x}^{2} - {2x}^{} + {1}^{2} ] = 0$

We know,

$\rm :\longmapsto\:\boxed{\tt{ {x}^{2} - 2xy + {y}^{2} = {(x - y)}^{2}}}$

So, using this identity, we get

$\rm :\longmapsto\: {( {x}^{2} - 1) }^{2} + x {(x - 1)}^{2} = 0$

We know,

$\rm :\longmapsto\:\boxed{\tt{ {x}^{2} - {y}^{2} = (x + y)(x - y)}}$

So, using this identity, we get

$\rm :\longmapsto\: {[(x - 1)(x + 1)] }^{2} + x {(x - 1)}^{2} = 0$

$\rm :\longmapsto\: {(x - 1)^{2} (x + 1) }^{2} + x {(x - 1)}^{2} = 0$

$\rm :\longmapsto\: {(x - 1)}^{2}[ {(x + 1)}^{2} + x] = 0$

$\rm :\longmapsto\: {(x - 1)}^{2}[ {x}^{2} + 1 + 2x + x] = 0$

$\rm :\longmapsto\: {(x - 1)}^{2}[ {x}^{2} + 3x + 1] = 0$

$\rm :\longmapsto\:x = 1,1 \: or \: {x}^{2} + 3x + 1 = 0$

Consider,

$\rm :\longmapsto \: {x}^{2} + 3x + 1 = 0$

To find the roots, we have to use Quadratic Formula.

So, using Quadratic formula, we get

$\rm :\longmapsto\:x = \dfrac{ - 3 \: \pm \: \sqrt{ {(3)}^{2} - 4(1)(1)} }{2(1)}$

$\rm :\longmapsto\:x = \dfrac{ - 3 \: \pm \: \sqrt{ 9 - 4} }{2}$

$\rm :\longmapsto\:x = \dfrac{ - 3 \: \pm \: \sqrt{5} }{2}$

So, Solution of

$\rm :\longmapsto\: {x}^{4} + {x}^{3} - {4x}^{2} + x + 1 = 0$

is

$\rm\implies \:\boxed{\tt{ x = 1, \: 1, \: \frac{ - 3 + \sqrt{5} }{2}, \: \frac{ - 3 - \sqrt{5} }{2}}}$

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Explore More :-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac

Answer 2

Step-by-step explanation:

x⁴ + x³ - 4x² + x + 1 = 0

Writes x³ as a sum.

x⁴ - x³ + 2x³ - 4x² + x + 1 = 0

Write -4x² and x as a sum.

x⁴ - x³ + 2x³ - + 1 = 0

Factor out: -

x³( x -1) + 2x³ - 2x² - 2x² + 2x - x + 1 = 0

x³( x -1) + 2x² ( x - 1) - 2x² + 2x - x + 1 = 0

x³( x -1) + 2x² ( x - 1) - 2x ( x - 1 ) - x + 1 = 0

x³( x -1) + 2x² ( x - 1) - 2x ( x - 1 ) - ( x - 1 ) = 0

Factor out (x - 1) from the expression.

( x - 1 ) ( x³ + 2x² - 2x - 1 ) = 0

Use identity :- a³ - b³ = ( a - ) ( a² + ab + b²).

( x - 1 ) ( ( x - 1 ) ( x² + x + 1 ) + 2x² - 2x ) = 0

Factor out 2x from the expression.

( x - 1) ( ( x - 1) ( x² + x + 1) + 2x ( x - 1)) = 0

Factor out ( x - 1) from the expression.

( x - 1) ( x - 1) ( x² + x + 1 + 2x) = 0

Collect like terms.

( x - 1) ( x - 1) ( x² + 3x + 1) = 0

Write the repeated multiplication in exponential form.

( x - 1) ² ( x ² + 3x - 1) = 0

When the products of factors equal zero, at least one factor is 0.

( x - 1 )² = 0

( x ² + 3x - 1) = 0

Solve for x.

x = 1

$\sf \small x \: = \frac{ - 3 + \sqrt{5} }{2} \:$

$\sf \small x \: = \frac{ - 3 \: - \sqrt{5} }{2} \:$

The equation has three solutions.

$x_1$= 1 , $x_2 = \sf \frac{ - 3 + \sqrt{5} }{2}$ and $\sf x_3 = \sf \frac{ - 3 + \sqrt{5} }{2}$