Math, asked by nissa8666, 1 year ago

Write the zeros of the quadratic equation f(x) =4√3x2+5x-2√3

Answers

Answered by Anonymous
5

hey mate

here is the answer...

Answer:

given :

4√3x²+5x-2√3=0

4√3x²+(8-3)x-2√3=0

4√3x²+8x-3x-2√3=0

4x(√3x+2) -3(√3x+2)=0

(4x-3) (√3x+2)=0

thus

x=3/4 or -2/√3

thus,

the zeros are 3/4 and -2/√3.

Answered by Anonymous
16

Answer :-

◾Provided Quadratic equation :-

 f(x) = 4\sqrt{3}x^2 + 5x - 2\sqrt{3}

▪️So , we have to find out the roots of the quadratic equation .

▪️We can use the Quadratic Formula to get the required Answer , which is :-

 x = \dfrac{-b \pm \sqrt{ b^2 - 4ac}}{2a}

So roots using Quadratic Formula :-

 = \dfrac{-(5) \pm\sqrt{(5)^2 - 4\times(4\sqrt{3})\times(-2\sqrt{3})}}{2\times 4\sqrt{3}}

 = \dfrac{-5 \pm \sqrt{ 25 - 4(-24)}}{8\sqrt{3}}

 = \dfrac{-5\pm\sqrt{ 25 + 96}}{8\sqrt{3}}

 = \dfrac{ -5 \pm \sqrt{ 121}}{8\sqrt{3}}

 = \dfrac{ -5 \pm 11}{8\sqrt{3}}

So :-

 x = \dfrac{ -5 + 11}{8\sqrt{3}}

 = \dfrac{ 6 }{8\sqrt{3}}

 = \dfrac{3}{4\sqrt{3}}

▪️And

 x = \dfrac{ -5 - 11}{8\sqrt{3}}

 = \dfrac{ -16}{8\sqrt{3}}

 = \dfrac{-2}{\sqrt{3}}

So roots of  \bold{x = \dfrac{3}{4\sqrt{3}} \: and \:\dfrac{-2}{\sqrt{3}}}

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