Question

solve for x
$\sqrt{2 {x}^{2} } + 7x + 5 \sqrt{2} = 0$
solve please fast​

Answer 1

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

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• $\frac{ - 5 \sqrt{2} }{2}$

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Given equation is

√2x² + 7x+5√/2

=

0

Now, splitting the middle terms, we get

√2x² + 2x + 5x+5√/2=0

can be further rewritten as

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

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

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

$x + \sqrt{2} = 0 \: or \: \sqrt{2x} + 5 = 0$

$x = - \sqrt{2} \: or \: \sqrt{2x } = - 5$

$x = - \sqrt{ 2} or \: x = \: - \frac{5}{ \sqrt{2} }$

$x = - \sqrt{2} or \: x = - \frac{5}{ \sqrt{2} } \times \frac{ \sqrt{2} }{ \sqrt{2} }$

$x = - \frac{5 \sqrt{2} }{ \sqrt{2} }$

Concept Used :

Splitting of middle terms :

In order to factorize ax² + bx + c we have

to find numbers m and n such that m + n

= b and mn = ac.

After finding m and n, we split the middle

term i.e bx in the given quadratic as mx +

nx and get required factors by grouping

the terms.

Additional Information :

Nature of roots :

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of

the equation are real and equal.

If Discriminant, D< 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac