Question

if one zeroes of polynomial x²-8x+k exceeds the other by 2, find the value of k

Answer 1

Hey mate, here is your answer:)

x^2-8x+k=0

on putting x= 2

(2)^2-8(2)+k=0

4-16+k=0

-8+k=0

k=8

Hence, the value of k is 8...

HOPE it helps!!!!

Answer 2

Answer:

Step-by-step explanation:

Concept:

Polynomial is made composed of the phrases Nominal, which suggests "terms," and Poly, which suggests "many." When exponents, constants, and variables are combined using mathematical operations like addition, subtraction, multiplication, and division, the result's a polynomial (No division operation by a variable). The classification of the phrase as a monomial, binomial, or trinomial depends on what percentage terms are included in it.

Solution:
Polynomial is $x^{2}-8 x+k$

Therefore, $a=1, b=-8, c=k$

Let one of the zeroes be $\alpha$

Therefore, other zero is $\alpha+2$

Sum of zeroes $\alpha+\alpha+2=-\frac{b}{a}$

                         $\begin{aligned}&= > 2 \alpha+2=-\frac{-8}{1} \\&= > 2 \alpha+2=8 \\&= > 2 \alpha=6 \\&= > \alpha=3\end{aligned}$

Therefore, one zero is 3

And other is $3+2=5$

Also, Product of zeroes $\alpha \times(\alpha+2)=\frac{c}{a}$

                        $\begin{aligned}&= > 3 \times 5=\frac{k}{1} \\&= > k=15\end{aligned}$

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