Question

find the zeroes of 4√3x² + 5x​

Answer 1

4√3x² + 5x

In the form of standard quadratic equation (ax² + bx + c), here,

a = 4√3

b = 5

c = 0

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So using the discriminant, we will find that zeroes are real or not.

discriminant = b² - 4ac

=> (5)² - 4(4√3)(0)

=> 25 - 0

=> 25

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Because discriminant (b² - 4ac) > 0, so the zeroes will be real and unequal.

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Now using the formula of zeroes of quadratic equation, we will find the zeroes

formula :-

$\frac{ - b ± ( \sqrt{ {b}^{2} - 4ac}}{2a} \\ = \frac{ - 5 ± \sqrt{25} }{2(4 \sqrt{3}) } \\ = \frac{ - 5 ± 5}{8 \sqrt{3} } \\ as \: - \\ \frac{ - 5 - 5}{8 \sqrt{3} } \\ = \frac{ - 10}{8 \sqrt{3} } \\ = \frac{ - 5}{4 \sqrt{3} } \\ \\ as + \\ \frac{ - 5 + 5}{8 \sqrt{3} } \\ = \frac{0}{8 \sqrt{3} } \\ = 0$

so zeroes = -5/4√3 and 0

hope it helps.