Question

The product of the real roots of the equation (x+1)4 +(x+3)4=8 is​

Answer 1

Given :  (x+1)⁴ +(x+3)⁴ = 8    

To find :   product of the real roots

Solution:

(x+1)⁴ +(x+3)⁴ = 8  

let say

x  = y - 2

=> y = x + 2

=> ( y - 1)⁴  + ( y + 1)⁴  = 8

=> 2y⁴  + 12y²  +  2 = 8

=> y⁴ + 6y²  + 1   = 4

=> y⁴ + 6y²  - 3  = 0

y² = z

=> z² + 6z  - 3 = 0

=> z = ( -6  ± √ (36 + 12) )/2

=> z  =  - 3 ± 2√3

z   = - 3 - 2√3   => y is not real  as y² is -ve

z =  - 3 + 2√3  

=> y²  =  2√3   - 3

y = x + 2

=> (x + 2)² = 2√3   - 3

=> x²  + 4x + 4  = 2√3   - 3

=> x²  + 4x + 7 - 2√3 = 0

as D > 0 hence roots are real

Products of roots  =   7 - 2√3

product of the real roots of the equation  (x+1)⁴ +(x+3)⁴ = 8  

is  7 - 2√3

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Answer 2

Step-by-step explanation:

Given:The product of the real roots of the equation (x+1)⁴ +(x+3)⁴=8 is

To find: Product of real roots

Solution:

To solve this

let

$y = x + 2$

So,

$({x + 1)}^{4} + ( {x + 3)}^{4} = 8 \\ \\ put \\ \\ y = x + 2 \\ \\ ({x + 2 - 1)}^{4} + ( {x + 2 + 1)}^{4} = 8 \\ \\( {y - 1)}^{4} + ( {y + 1)}^{4} = 8$

Expand the equation in y

$( {y}^{2} + 1 - 2y)( {y}^{2} + 1 - 2y) + ( {y}^{2} + 1 + 2y)( {y}^{2} + 1 + 2y) = 8 \\ \\ {y}^{4} + {y}^{2} - 2 {y}^{3} + {y}^{2} + 1 - 2y - 2 {y}^{3} - 2y + 4 {y}^{2} + {y}^{4} + {y}^{2} + 2 {y}^{3} + {y}^{2} + 1 + 2y + 2 {y}^{3} + 2y + 4 {y}^{2} = 8 \\ \\ 2 {y}^{4} + 12 {y}^{2} + 2 = 8 \\ \\ {y}^{4} + 6 {y}^{2} - 3 = 0$

its a quadratic equation in y²

solve this with the help of quadratic formula

${y}^{4} + 6 {y}^{2} - 3 = 0 \\ {y}^{2}=\frac{-b±\sqrt{b^2-4ac}}{2a}\\\\ {y}^{2}=\frac{-6±\sqrt{36+12}}{2}\\\\ {y}^{2}=\frac{-6±\sqrt{48}}{2}\\\\{y}^{2}=\frac{-6±4\sqrt{3}}{2}\\\\{y}^{2} = -3±2\sqrt{3}$

Value of y= -3-2√3,gives imaginary roots

So, real roots can be calculated easily

${y}^{2} = -3+2\sqrt{3}$

Now put the value of y in substitution

Case1: y= 0.68

$y = x + 2 \\ \\ 0.68= x + 2 \\ \\ x = 1.32$

Case2:y=-0.68

$y = x + 2 \\ \\ -0.68= x + 2 \\ \\ x = 2.68$

Product of real roots are:(1.32)×(2.68) =3.54

Hope it helps you.