The product of the real roots of the equation (x+1)^4 +(x+3)^4=8 is 2
Answers
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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Step-by-step explanation:
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
Learn more:
One of the roots of the quadratic equation 4mnx2 – 6m2x – 6n2x + ...
https://brainly.in/question/18994810
if the product of two roots of the equation 4x^4-24x^3+31x^2+6x-8=0 ...
https://brainly.in/question/18325992