Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and the coefficients.
$4 x^{2} - 3 - 7 x$

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Solution

We have,

$4 x^{2} - 3 x - 7$

$= 4 x^{2} - 7 x + 4 x - 7$

$= x (4 x - 7) + 1 (4 x - 7)$

$= (4 x - 7) (x + 1)$

So, the value of $4 x^{2} - 3 x - 7$ is zero when $4 x - 7 = 0$ or $x + 1 = 0$

i.e., when $x = \frac{7}{4}$ or $x = - 1$

Therefore, the zeroes of $4 x^{2} - 3 x - 7$ are $\frac{7}{4}$ and $- 1$

Now, Sum of zeroes $= \frac{7}{4} + (- 1) = \frac{3}{4} = \frac{- (- 3)}{4} = \frac{- b}{a}$

Product of zeroes $= \frac{7}{4} \times (- 1) = \frac{- 7}{4} = \frac{c}{a}$

Hence verified.

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