Question

$8x { }^{2} + 2 \sqrt{ 3} x - 9 = 0$
solve​

Answer 1

Question:-

$8 {x}^{2} + 2 \sqrt{3} x - 9 = 0$

Solution:-

We have equation:-

$8 {x}^{2} + 2 \sqrt{3} x - 9 = 0$

Compare with:- ax² + bx² + c = 0

We get ,

$\rm \: a \: = \: 8 \: \: or \: \: b \: = 2 \sqrt{3} \: \: or \: \: \: c \: = \: - 9$

Using the quadratic formula

$\rm \: x = \frac{ - b \pm \sqrt{D} }{2a}$

$\rm \: D = {b}^{2} - 4ac$

$\rm \: D = (2 \sqrt{3} ) {}^{2} - 4 \times 8 \times - 9$

$\rm \: D = 12 + 288$

$\rm \: D = 300$

Using this formula

$\rm \: x = \frac{ - b \pm \sqrt{D} }{2a}$

We get

$\rm \: x = \frac{ - 2 \sqrt{3} \pm \sqrt{300} }{2 \times 8}$

$\rm \: x = \frac{ - 2 \sqrt{3} \pm10 \sqrt{3} }{16}$

$\rm \: x \: = \frac{ - 2 \sqrt{3} + 10 \sqrt{3} }{16} \: \: \: and \: \: \: x = \frac{ - 2 \sqrt{ 3} - 10 \sqrt{3} }{16}$

$\rm \: x = \frac{8 \sqrt{3} }{16} \: \: and \: \: \: x = \frac{ - 12 \sqrt{3} }{16}$

$\color{red} \rm \: x = \frac{ \sqrt{3} }{2} \: \: \: and \: \: \:x = \frac{ - 3 \sqrt{3} }{4}$

Answer 2

$\sf{\underline{\boxed{\green{\large{\bold{ Question}}}}}}$

• solve the equation using formulae $\sf 8x^2 + 2\sqrt {3x} - 9 = 0$

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$\sf{\underline{\boxed{\green{\large{\bold{ Solution}}}}}}$

$\sf\implies 8x^2 + 2 \sqrt{3x} - 9 = 0$

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• compare the eq with $\sf{\underline{\bold{ax^2 + bx + c = 0 }}}$

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☯ a = 8

☯ b = 2√3

☯ c = -9

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now :-

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$\sf{\underline{\boxed{\pink{\large{\mathfrak{x = \dfrac{ - b \pm \sqrt D }{2a }}}}}}}$

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$\sf{\underline{\boxed{\pink{\large{\mathfrak{ D = b^2 - 4ac }}}}}}$

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• finding value of D.

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$\sf\implies D = b^2 - 4ac$

$\sf\implies D = (2\sqrt 3)^2 - 4 \times 8 \times -9$

$\sf\implies D = 12 + 288$

$\sf\implies D = 300$

$\sf{\underline{\boxed{\blue{\large{\bold{ D = 300}}}}}}$

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• putting values in the eq.

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$\sf\implies x = \dfrac{ -b \pm\sqrt D }{2a}$

$\sf\implies x = \dfrac{ -( 2\sqrt 3) \pm\sqrt {300} }{2\times 8 }$

$\sf\implies x = \dfrac{ -( 2\sqrt 3) \pm 10\sqrt {3} }{2\times 8 }$

$\sf\implies x = \dfrac{ -2\sqrt 3 \pm 10\sqrt 3 }{16}$

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✒$\sf x = \dfrac{ -2 \sqrt 3 + 10\sqrt 3}{ 16 }$

$\implies x = \dfrac {8\sqrt 3}{16}$

$\implies x = \dfrac{\sqrt 3}{2}$

$\sf{\underline{\boxed{\purple{\large{\bold{ x = \dfrac{\sqrt 3}{2} }}}}}}$

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✒$\sf x = \dfrac{ -2 \sqrt 3 - 10\sqrt 3}{ 16 }$

$\implies x = \dfrac {-12\sqrt 3}{16}$

$\implies x = \dfrac{-3\sqrt 3}{4}$

$\sf{\underline{\boxed{\purple{\large{\bold{ x = \dfrac{-3\sqrt 3}{4} }}}}}}$

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$\sf{\underline{\boxed{\purple{\large{\bold{ x = \dfrac{\sqrt 3}{2} \: or \: \dfrac{-3\sqrt 3}{4}}}}}}}$