Math, asked by svkenterprises, 6 months ago

x^2-2x+8 find zeroes​

Answers

Answered by neerubhola83
2

Answer:

Other zeroes are : x = 4 , x = 2

Step-by-step explanation:

HOPE IT WILL HELP:-)))

Attachments:
Answered by Anonymous
7

Answer:

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.

I think it will help you

Attachments:
Similar questions