Question

4√3x square +5x-2√3 using quadratic formula​

Answer 1

Answer:

4√3 x² +5x -2√3 =0

⇒4√3 x² +8x -3x -2√3=0

⇒4x(√3 x+2) -√3(√3x +2) =0

⇒(√3x +2)(4x -√3) =0

⇒(√3x +2) =0 or (4x-√3) =0

⇒x = -2/√3 or √3/4

Step-by-step explanation:

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Answer 2

Answer:

$\huge\boxed{x=-\dfrac{2\sqrt3}{3}\ \vee\ x=\dfrac{\sqrt3}{4}}\\4\sqrt3x^2+5x-2\sqrt3=4\sqrt3\left(x+\dfrac{2\sqrt3}{3}\right)\left(x-\dfrac{\sqrt3}{4}\right)$

Step-by-step explanation:

The quadratic formula:

$ax^2+bx+c=0\\\\\text{if}\ b^2-4ac<0,\ \text{then no solution}\\\text{if}\ b^2-4ac=0,\ \text{then one solution}:\ x=\dfrac{-b}{2a}\\\text{if}\ b^2-4ac>0,\ \text{then two solutions}:\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

We have

$4\sqrt3x^2+5x-2\sqrt3=0\\\\a=4\sqrt3,\ b=5,\ c=-2\sqrt3\\\\b^2-4ac=5^2-4\cdot4\sqrt3\cdot(-2\sqrt3)=25+96=121>0\\\\\sqrt{b^2-4ac}=\sqrt{121}=11\\\\x_1=\dfrac{-5-11}{2(4\sqrt3)}=\dfrac{-16}{8\sqrt3}=\dfrac{-2}{\sqrt3}\cdot\dfrac{\sqrt3}{\sqrt3}=-\dfrac{2\sqrt3}{3}\\\\x_2=\dfrac{-5+11}{2(4\sqrt3)}=\dfrac{6}{8\sqrt3}=\dfrac{3}{4\sqrt3}\cdot\dfrac{\sqrt3}{\sqrt3}=\dfrac{3\sqrt3}{4\cdot3}=\dfrac{\sqrt3}{4}$