Question

Please help quadratic equations

Answer 1

Question:-

$\to \: \rm \: eq \: \: : \: 2 \sqrt{3} {x}^{2} + 7x + 2 \sqrt{3} = 0$

Solution:-

$\to \: \rm \: eq \: \: : \: 2 \sqrt{3} {x}^{2} + 7x + 2 \sqrt{3} = 0$

Compare with :- ax² + bx + c = 0 , we get

$\rm \: a = 2 \sqrt{3} \: \: \: \: b \: = \: 7 \: \: and \: \: c \: = 2 \sqrt{ 3}$

Using quadratic formula

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

Find discriminant ( D )

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

We get

$\rm \: D = (7) {}^{2} - 4 \times 2 \sqrt{3} \times 2 \sqrt{3}$

$\rm \: D = 49 - 4 \times 12$

$\rm \: D = 49 - 48$

$\rm \: D = 1$

Now put the value in quadratic formula we get

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

$\rm \: x = \frac{ - 7 \pm \sqrt{1} }{2 \times 2 \sqrt{3} }$

$\rm \: x = \frac{ - 7 + 1}{2 \sqrt{3} } \: \: and \: \: \frac{ - 7 - 1}{2 \sqrt{3} }$

$\rm \: x = \frac{ - 6}{2 \sqrt{3} } \: \: and \: \: \frac{ - 8}{2 \sqrt{3} }$

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

Now rationalize the denominator, we get

$\rm \: x = \frac{ - 3}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } \: \: and \: \frac{ - 4}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} }$

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

So , final value of x is

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

Answer 2

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

• solve the equation using formulae $\sf 2\sqrt 3 x^2 + 7x + 2\sqrt 3 = 0$

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

$\sf\implies 2\sqrt 3 x^2 + 7x + 2\sqrt 3= 0$

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

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☯ a = 2√3

☯ b = 7

☯ c = 2√3

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

$\sf\implies D = (7)^2 - 4 \times 2\sqrt 3\times 2\sqrt 3$

$\sf\implies D = 49 - 48$

$\sf\implies D = 1$

$\sf{\underline{\boxed{\blue{\large{\bold{ D = 1}}}}}}$

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• putting values in the eq.

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

$\sf\implies x = \dfrac{ -( 7 ) \pm\sqrt {1} }{2\times 2\sqrt 3}$

$\sf\implies x = \dfrac{ -7 \pm 1 }{4\sqrt 3}$

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✒$\sf x = \dfrac{ -7 + 1 }{ 4\sqrt 3 }$

$\implies x = \dfrac {-6}{4\sqrt 3}$

• Rationalize the denominator

$\implies x = \dfrac {-6}{4\sqrt 3}\times\dfrac{4\sqrt 3}{4\sqrt 3}$

$\implies x = \dfrac {-24\sqrt 3}{48}$

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

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

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

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✒$\sf x = \dfrac{ -7 - 1 }{ 4\sqrt 3 }$

$\implies x = \dfrac {-8}{4\sqrt 3}$

Rationalize the denominator

$\implies x = \dfrac {-8}{4\sqrt 3}\times\dfrac{4\sqrt 3}{4\sqrt 3}$

$\implies x = \dfrac {-32\sqrt 3}{48}$

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

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

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

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$\sf{\underline{\boxed{\purple{\large{\bold{ x = \dfrac{-\sqrt 3}{2} \: or \:-\dfrac{-2\sqrt3}{3}}}}}}}$