Question

Q:-solve the quadratic equation
$\sqrt{2} {x}^{2} + 7x + 5 \sqrt{2} = 0$ find value of x using factorisation method​

Answer 1

$\green{\bold{\underline{ ☆ UPSC-ASPIRANT ☆} }}$

$\red{\bold{\underline{\underline{QUESTION:-}}}}$

Q:-solve the quadratic equation

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

$\huge\tt\underline\blue{ANSWER }$

------>>>>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..

_____________________

Thankyou:)

Answer 2

Answer:

⇛2x2+7x+52=0

⇛ \sqrt{2} {x}^{2} + 2x + 5x + 5 \sqrt{2} = 0⇛2x2+2x+5x+52=0

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

⇛(x + \sqrt{2} )( \sqrt{2} x + 5)⇛(x+2)(2x+5)

⇛x + \sqrt{2} = 0⇛x+2=0

⇛x = - \sqrt{2} ✓⇛x=−2✓

⇛ \sqrt{2} x + 5 = 0⇛2x+5=0

⇛x = - \frac{5}{ \sqrt{2} } ✓⇛x=−25✓