Let a be the largest real root and b be the s | vedclass

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

Question

(B) The given polynomial equation is $x^6-6x^5+15x^4-20x^3+15x^2-6x+1=0$. This expression follows the binomial expansion of $(x-1)^6$. Thus,the equati

Vedclass · Accepted Answer

$\frac{2}{3}$

Vedclass · Answer

(B) The given polynomial equation is $x^6-6x^5+15x^4-20x^3+15x^2-6x+1=0$. This expression follows the binomial expansion of $(x-1)^6$. Thus,the equation can be written as $(x-1)^6=0$. This implies that all roots of the equation are equal to $1$. Therefore,the largest real root $a = 1$ and the smallest real root $b = 1$. Substituting these values into the expression,we get $\frac{a^2+b^2}{a+b+1} = \frac{1^2+1^2}{1+1+1} = \frac{1+1}{3} = \frac{2}{3}$.

Let $a$ be the largest real root and $b$ be the smallest real root of the polynomial equation $x^6-6x^5+15x^4-20x^3+15x^2-6x+1=0$. Then $\frac{a^2+b^2}{a+b+1}$ is

Similar Questions

The number of terms in the expansion of $(x^{2}+y^{2})^{25}-(x^{2}-y^{2})^{25}$ after simplification is

If $1 + x^4 + x^5 = \sum\limits_{i = 0}^5 a_i (1 + x)^i$ for all $x$ in $\mathbb{R},$ then $a_2$ is

If $\sum_{r=0}^{10} \left( \frac{10^{r+1}-1}{10^r} \right) \cdot {}^{11}C_{r+1} = \frac{\alpha^{11}-11^{11}}{10^{10}}$,then $\alpha$ is equal to :

Expand the expression $\left(\frac{x}{3}+\frac{1}{x}\right)^{5}$

If $a_k$ is the coefficient of $x^k$ in the expansion of $(1+x+x^2)^n$ for $k=0, 1, 2, \ldots, 2n$,then $a_1+2a_2+3a_3+\ldots+2na_{2n}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module