The product of the real roots of the equation | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given equation: $(x+1)^4+(x+3)^4=8$. Let $x+2=y$,then $x+1=y-1$ and $x+3=y+1$. The equation becomes $(y-1)^4+(y+1)^4=8$. Expanding using binomial

Vedclass · Accepted Answer

$7-2 \sqrt{3}$

Vedclass · Answer

(C) Given equation: $(x+1)^4+(x+3)^4=8$. Let $x+2=y$,then $x+1=y-1$ and $x+3=y+1$. The equation becomes $(y-1)^4+(y+1)^4=8$. Expanding using binomial theorem: $(y^4-4y^3+6y^2-4y+1)+(y^4+4y^3+6y^2+4y+1)=8$. $2y^4+12y^2+2=8$ $\Rightarrow 2y^4+12y^2-6=0$ $\Rightarrow y^4+6y^2-3=0$. Let $t=y^2$,then $t^2+6t-3=0$. $t = \frac{-6 \pm \sqrt{36-4(1)(-3)}}{2} = \frac{-6 \pm \sqrt{48}}{2} = -3 \pm 2\sqrt{3}$. Since $t=y^2$ must be non-negative for real $y$,we take $y^2 = 2\sqrt{3}-3$. Substituting back $y=x+2$: $(x+2)^2 = 2\sqrt{3}-3$. $x^2+4x+4 = 2\sqrt{3}-3 \Rightarrow x^2+4x+(7-2\sqrt{3}) = 0$. This is a quadratic equation in $x$ of the form $ax^2+bx+c=0$. The product of the roots is $\frac{c}{a} = \frac{7-2\sqrt{3}}{1} = 7-2\sqrt{3}$.

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

Similar Questions

If $x$ is real,then the interval in which no value of the expression $\frac{2(x^2+2x-11)}{2x-5}$ lies,is

If $\frac{k}{kx+3}+\frac{3}{3x-k}=\frac{12x+5}{(kx+3)(3x-k)}$ for all $x \in R - \{-\frac{3}{k}, \frac{k}{3}\}$,then both the roots of the equation $kx^2-7x+3=0$ are

The set of all values of $k > -1$,for which the equation $(3x^2 + 4x + 3)^2 - (k + 1)(3x^2 + 4x + 3)(3x^2 + 4x + 2) + k(3x^2 + 4x + 2)^2 = 0$ has real roots is:

The algebraic equation of degree $4$ whose roots are the translates of the roots of the equation $x^4+5x^3+6x^2+7x+9=0$ by $-1$ is

Sum of the solutions of the equation $[x^2] - 2x + 1 = 0$ is (where $[.]$ denotes the greatest integer function).

Vedclass Test Series

Exam Paper Generator

Online Exam Module