Question

find the zeros of the quadratic polynomial x^2 + 7/2x +3/4 and verify the relationship between the zeros and the coefficients

Answer 1

Heya mate,
here is your answer,
_____________

We have
$p(x) = {x}^{2} + \frac{7x}{2} + \frac{3}{4}$

Now, zeroes of p(x) is given by ,
p (x) = 0
$= > {x}^{2} + \frac{7x}{2} + \frac{3}{4} = 0 \\ = > {4x}^{2} + 14x + 3 = 0$
Now, a = 4 ; b = 14 ; c = 3

D = b^2 - 4ac
$= > {14}^{2} - 4(4)(3) = 196 - 48 = 148 > 0$

So, by quadratic formula,
$x = \frac{ - b + \sqrt{d} }{2a} \: and \: \: x = \frac{ - b - \sqrt{d} }{2a}$

$= > x = \frac{ - 14 + \sqrt{148} }{8} \: \: and \: \: x = \frac{ - 14 - \sqrt{148} }{8}$

$= > x = \frac{ - 14 + 2 \sqrt{37} }{8} \: and \: \: x = \frac{ - 14 - 2 \sqrt{37} }{8} \\ \\ = > x = \frac{ - 7 + \sqrt{37} }{4} \: \: and \: \: x = \frac{ - 7 - \sqrt{37} }{4}$

Now , sum of zeroes =
$\frac{ - 7 + \sqrt{37} }{4} + \frac{ - 7 - \sqrt{37} }{4} = \frac{ - 7}{2} = - \frac{coefficient \: of \: x}{coefficient \: of \: {x}^{2} }$

product of zeroes =
$\frac{ - 7 + \sqrt{37} }{4} \times \frac{ - 7 - \sqrt{37} }{4} = \frac{49 - 37}{16} = \frac{12}{16} = \frac{3}{4} = \frac{constant \: term}{coefficient \: of \: {x}^{2} }$

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