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Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and their coefficients:
$f(x)\ =\ x^2\ β\ 2x\ β\ 8$
$f(x)\ =\ x^2\ β\ 2x\ β\ 8$
Given:
$f(x) = x^2 – 2x – 8$
To find:
Here, we have to find the zeros of f(x).
Solution:
To find the zeros of f(x), we have to put $f(x)=0$.
This implies,
$x^2 – 2x – 8 = 0$
$x^2 – 4x + 2x – 8 = 0$
$x(x – 4) + 2(x – 4) = 0$
$(x – 4)(x + 2) = 0$
$x-4=0$ and $x+2=0$
$x = 4$ and $x = -2$
Therefore, the zeros of the quadratic equation $f(x) = x^2 – 2x – 8$ are $4$ and $-2$.
Verification:
We know that,
Sum of zeros $= -\frac{coefficient of x}{coefficient of x^2}$
$= –\frac{(-2)}{1}$
$=2$
Sum of the zeros of $f(x)=4+(-2)=4-2=2$
Product of roots $= \frac{constant}{coefficient of x^2}$
$= \frac{(-8)}{1}$
$= -8$
Product of the roots of $f(x)=4\times(-2) =-8$
Hence, the relationship between the zeros and their coefficients is verified.
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