Question

8x^2 +5√2x + 1 =0 find factors​

Answer 1

the answer is this question them me solve this question

the questions is answer is 64 is answer

Answer 2

Basic Concept Used :-

There are three methods to find the factors of quadratic equation :-

• 1. Method of factorization

• 2. Method of Completing squares

• 3. Using Quadratic Formula

Here, we prefer

Method of Factorization using splitting of middle terms.

Splitting of middle terms :-

• In order to factorize  ax² + bx + c we have to find numbers p and q such that p + q = b and pq = ac.

• After finding p and q, we split the middle term in the quadratic as px + qx and get desired factors by grouping the terms.

❥ Solution :-

$\rm :\longmapsto\: {8x}^{2} + 5 \sqrt{2}x + 1 = 0$

Step :- 1 We have to find p and q such that

$\red{\rm :\longmapsto\:pq = 8 \: and \: p + q = 5 \sqrt{2}}$

$\red{\rm :\longmapsto\:pq = 4 \sqrt{2} \times \sqrt{2} \: and \: p + q = 4 \sqrt{2} + \sqrt{2} }$

$\red{\rm :\longmapsto\:\boxed{\bf \:p = 4 \sqrt{2} \: and \: q = \sqrt{2}}}$

Step :- 2

$\rm :\longmapsto\: {8x}^{2} + (4 \sqrt{2} + \sqrt{2})x + 1 = 0$

$\rm :\longmapsto\: {8x}^{2} + 4 \sqrt{2}x + \sqrt{2}x + 1 = 0$

$\rm :\longmapsto\: {4\times\sqrt{2} \times\sqrt{2}\times x}^{2}+4\sqrt{2}x +\sqrt{2}x +1 = 0$

$\rm :\longmapsto\:4 \sqrt{2}x( \sqrt{2}x + 1) + 1( \sqrt{2}x + 1) = 0$

$\rm :\longmapsto\:(4 \sqrt{2}x + 1)( \sqrt{2}x + 1) = 0$

$\rm :\longmapsto\:4\sqrt{2}x +1=0 \: \: or \: \: \sqrt{2}x + 1 = 0$

$\rm :\longmapsto\:x = - \dfrac{1}{4 \sqrt{2}} \: \: \: or \: x = - \dfrac{1}{ \sqrt{2}}$

$\rm :\longmapsto\:x = - \dfrac{1}{4 \sqrt{2}} \times \dfrac{ \sqrt{2} }{ \sqrt{2} } \: \: \: or \: x = - \dfrac{1}{ \sqrt{2}} \times \dfrac{ \sqrt{2} }{ \sqrt{2} }$

$\rm :\longmapsto\:x = - \dfrac{ \sqrt{2} }{8} \: \: \: or \: x = - \dfrac{ \sqrt{2} }{2}$

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