F(x)=x2-2x-8find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficient.
Answers
Given : f(x)=x²-2x-8
To Find : Zeroes
Solution :
f(x)=x²-2x-8
using middle term spilt
x²-2x-8
= x² - 4x + 2x - 8
= x(x - 4) + 2(x - 4)
= (x - 4)(x + 2)
to Find zeroes
(x - 4)(x + 2) = 0
=> x = 4 , x = - 2
Zeroes are 4 , - 2
ax² + bx + c = 0
=> Sum of zeroes = - b/a
Product of zeroes = c/a
a = 1 , b = - 2 , c = - 8
Zeroes are 4 , - 2
Sum of Zeroes = 2 = -(-2)/1 = 2
Products of Zeroes = - 8 = (-8)/1 = -8
Verified
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Given,
f(x) = x2 – 2x – 8
To find the zeros, we put f(x) = 0
⇒ x
2 – 2x – 8 = 0
⇒ x
2
- 4x + 2x - 8 = 0
⇒ x(x - 4) + 2(x - 4) = 0
⇒ (x - 4)(x + 2) = 0
This gives us 2 zeros, for
x = 4 and x = -2
Hence, the zeros of the quadratic equation are 4 and -2.
Now, for verification
Sum of zeros = - coefficient of x / coefficient of x2
4 + (-2)= - (-2) / 1
2 = 2
Product of roots = constant / coefficient of x2
4 x (-2) = (-8) / 1
-8 = -8
Therefore, the relationship between zeros and their coefficients is verified.