Question

$\sqrt{2 {x}^{2} } + 7x + 5 \sqrt{2 } = 0$
please solve please... quickly . I will mark you as brainlist​

Answer 1

Step-by-step explanation:

2 x 2+7x+5 2=0

or, 2x 2+2x+5x+5 2=0

or, x 2(x+ 2)+5(x+ 2)=0

or, (x 2+5)(x+2)=0

⇒x= 25 ,−2.

These are the required roots.

Answer 2

Appropriate Question :- Solve for x :-

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

$\large\underline{\sf{Solution-}}$

Given equation is

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

Now, splitting the middle terms, we get

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

can be further rewritten as

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

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

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

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

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

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

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

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

$\rule{190pt}{2pt}$

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.

$\rule{190pt}{2pt}$

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