find the zeroes of the polynomial x2-2x-8
Answers
Answer and Explanation:
Given : Quadratic polynomial x^2-2x-8x
2
−2x−8
To find : The zeros of the quadratic polynomial relationship between zeroes and coefficients ?
Solution :
First we solve the quadratic polynomial to get the roots of the polynomial.
Applying Middle term split,
x^2-2x-8=0x
2
−2x−8=0
x^2-4x+2x-8=0x
2
−4x+2x−8=0
x(x-4)+2(x-4)=0x(x−4)+2(x−4)=0
(x-4)(x+2)=0(x−4)(x+2)=0
(x-4)=0,(x+2)=0(x−4)=0,(x+2)=0
x=4,x=-2x=4,x=−2
So, The roots of the quadratic polynomial are \alpha=4,\beta=-2α=4,β=−2
The zeros of the polynomial are
\begin{gathered}\alpha+\beta=4-2=2\\\alpha \beta=4(-2)=-8\end{gathered}
α+β=4−2=2
αβ=4(−2)=−8
The zeros of the quadratic polynomial relationship between zeroes and coefficients is
Let a is the coefficient of x², b is the coefficient of x and c is the constant
i.e. Substituting, a=1,b=-2 and c=-8
Sum of zeros is
\alpha+\beta=-\frac{b}{a}α+β=−
a
b
\alpha+\beta=-\frac{-2}{1}α+β=−
1
−2
\alpha+\beta=2α+β=2
It is verified.
Product of zeros is
\alpha\beta=\frac{c}{a}αβ=
a
c
\alpha\beta=\frac{-8}{1}αβ=
1
−8
\alpha\beta=-8αβ=−8
It is verified.
Answer:
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