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Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and their coefficients:
$f(x)\ =\ 6x^2\ β\ 3\ β\ 7x$
Given:
$f(x) = 6x^2 – 3-7x$
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,
$6x^2 – 3-7x = 0$
$6x^2 –7x-3 = 0$
$6x^2 – 9x +2x -3 = 0$
$3x(2x – 3) +1(2x– 3) = 0$
$(2x– 3)(3x+1) = 0$
$2x-3=0$ and $3x+1=0$
$2x= 3$ and $3x= -1$
$x=\frac{3}{2}$ and $x=\frac{-1}{3}$
Therefore, the zeros of the quadratic equation $f(x) = 6x^2 – 3 -7x$ are $\frac{3}{2}$ and $\frac{-1}{3}$.
Verification:
We know that,
Sum of zeros $= -\frac{coefficient of x}{coefficient of x^2}$
$= –\frac{(-7)}{6}$
$=\frac{7}{6}$
Sum of the zeros of $f(x)=\frac{3}{2}+\frac{-1}{3}=\frac{3\times3-1\times2}{6}=\frac{9-2}{6}=\frac{7}{6}$
Product of roots $= \frac{constant}{coefficient of x^2}$
$= \frac{-3}{6}$
$=\frac{-1}{2}$
Product of the roots of $f(x)=\frac{3}{2}\times\frac{-1}{3} =\frac{-1}{2}$
Hence, the relationship between the zeros and their coefficients is verified.
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