Question

Q:-solve the quadratic equation with proper steps
$\sqrt{2} {x}^{2} + 7x + 5 \sqrt{2} = 0$​

Answer 1

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Q:-solve the quadratic equation

$\sqrt{2} {x}^{2} + 7x + 5 \sqrt{2} = 0$

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------>>>>Here is your answer<<<<--------

$⇛ \sqrt{2} {x}^{2} + 7x + 5 \sqrt{2} = 0$

$⇛ \sqrt{2} {x}^{2} + 2x + 5x + 5 \sqrt{2} = 0$

$⇛ \sqrt{2} x(x + \sqrt{2} ) + 5(x + \sqrt{2} ) = 0$

$⇛(x + \sqrt{2} )( \sqrt{2} x + 5)$

$⇛x + \sqrt{2} = 0$

$⇛x = - \sqrt{2} ✓$

$⇛ \sqrt{2} x + 5 = 0$

$⇛x = - \frac{5}{ \sqrt{2} } ✓$

HOPE IT HELPS YOU..

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Thankyou:)

Answer 2

Given:-

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$\bf \implies \sqrt{2} {x}^{2} + 7x + 5 \sqrt{2} = 0$

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To Find:-

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$\bf \implies the \: solution$

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STEP BY STEP EXPLANATION:-

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$\bf \implies \sqrt{2} {x}^{2} + 7x + 5 \sqrt{2} = 0 \\ \\ \bf \implies x = \frac{ - 7± \sqrt{ {7}^{2} - 4 \sqrt{2 } \times 5 \sqrt{2} } }{2 \sqrt{2} } \\ \\ \bf \implies\frac{ - 7± \sqrt{ 49 - 4 \sqrt{2 } \times 5 \sqrt{2} } }{2 \sqrt{2} } \\ \\ \bf \implies\frac{ - 7± \sqrt{ 49 - 4 \times 2 \times 5 \sqrt{2} } }{2 \sqrt{2} } \\ \\ \bf \implies\frac{ - 7± \sqrt{ 49 - 4 0 } }{2 \sqrt{2} } \\ \\ \bf \implies\frac{ - 7± \sqrt{ 9} }{2 \sqrt{2} } \\ \\ \bf \implies\frac{ - 7± { 3 }}{2 \sqrt{2} } \\ \\ \bf \implies\frac{ - 7 + { 3 }}{2 \sqrt{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf \frac{ - 7 - { 3 }}{2 \sqrt{2} } \\ \\ \bf \implies x = - \sqrt{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf \frac{ - 7 - { 3 }}{2 \sqrt{2} } \\ \\ \bf \implies x = - \sqrt{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf \frac{ 5 \sqrt{2} }{2 } \\ \\ \bf \implies x_1 = - \frac{5 \sqrt{2} }{2} \: \: \: \: \: , \: \: \: \: \: x_2 = - \sqrt{2}$