Question

Find the zeroes of the polynomial x2 -x-72 and verify the relation between the zeroes and the coefficients.

Answer 1

so A.T.Q
x²-x-72
x²-9x+8x-72
x(x-9)+8(x-9)
(x-9)(x+8)
x=9:x=(-8)
let a =9 and ß=(-8)
so aß=c/a
9(-8)=-72
a+ß=-b/a
9-8=-(-1)
1=1
hence 9 and -8 verify the relation between the zeroes and the coefficients.
I hope it would be the brainiest one

Answer 2

Answer:

Concept:

All the x-values that bring a polynomial, p(x), to zero are referred to as its zeros. They are intriguing to us for a variety of reasons, one of which is that they provide information on the graph's x-intercepts for the polynomial. We will also observe that they have a direct connection to the polynomial factors.

Step-by-step explanation:

Given:

The polynomial $x^{2}$-x-72

Find:

Find the zeroes of the polynomial $x^{2}$-x-72 and verify the relation between the zeroes and the coefficients.

Solution:

          $\begin{aligned}&p(x)=x^{2} -x-72=0 \\&=x^{2} -9 x+8 x-72=0 \\&=x(x-9)+8(x-9)=0 \\&=(x-9)+(x+8)=0 \\&x-9=0 \ x+8=0 \\&x=9 \ x=-8\end{aligned}$

     On comparing $a x^{2} +b x+c=0$

         We get,

         a = 1, b = -1, c = -72

      Product of zeroes

      Alpha × beta = c/a

          $\begin{aligned}&9 \times-8=-72 / 1 \\&-72=-72\end{aligned}$

       Sum of zeroes

       Alpha + beta= -b/a

           $\begin{aligned}&9+(-8)=-(-1 / 1) \\&9-8=1 / 1 \\&1=1\end{aligned}$

    Hence, the relationship is verified.

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