find the zeros of quadratic polynomial x^2-2x-8, and verify the relationship between the eros and coefficients.
Answers
P(X) => X²-2X-8
=> X²-4X+2X-8
=> X(X-4) +2( X-4)
=> (X-4) (X+2) = 0
=> (X-4) = 0 OR (X+2) = 0
=> X = 4 OR X = -2
4 and -2 are the two zeroes of the polynomial X²-2X-8.
RELATIONSHIP BETWEEN THE ZEROES AND COEFFICIENT.
Sum of zeroes = Alpha + Beta = 4 + (-2) = 4-2 = 2/1 = Coefficient of X/Coefficient of X².
And,
Product of zeroes = Alpha × Beta = 4 × -2 = -8/1 = Constant term/Coefficient of X².
HOPE IT WILL HELP YOU...... :-)
Step-by-step explanation:
First we solve the quadratic polynomial to get the roots of the polynomial.
Applying Middle term split,
x^2-2x-8=0x2−2x−8=0
x^2-4x+2x-8=0x2−4x+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
$$\begin{lgathered}\alpha+\beta=4-2=2\\\alpha \beta=4(-2)=-8\end{lgathered}$$
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
$$\alpha+\beta=-\frac{b}{a}$$
$$\alpha+\beta=-\frac{-2}{1}$$
$$\alpha+\beta=2$$
It is verified.
Product of zeros is
$$\alpha\beta=\frac{c}{a}$$
$$\alpha\beta=\frac{-8}{1}$$
$$\alpha\beta=-8$$
It is verified.