Question

find the roots of 6x^-root 2x - 2 = 0​

Answer 1

Appropriate Question :- Find thr roots of

$\rm \: {6x}^{2} - \sqrt{2} x - 2 = 0$

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

Given equation is

$\rm \: {6x}^{2} - \sqrt{2} x - 2 = 0$

On splitting the middle terms, we get

$\rm \: {6x}^{2} - 3\sqrt{2} x + 2 \sqrt{2}x - 2 = 0$

can be further rewritten as

$\rm \: 3. \sqrt{2}. \sqrt{2} {x}^{2} - 3\sqrt{2} x + 2 \sqrt{2}x - 2= 0$

$\rm \: 3 \sqrt{2}x( \sqrt{2}x - 1) + 2( \sqrt{2}x - 1) = 0$

$\rm \: ( \sqrt{2}x - 1)( 3\sqrt{2}x + 2) = 0$

$\rm \: \sqrt{2}x - 1= 0 \: \: or \: \: 3\sqrt{2}x + 2 = 0$

$\rm \: \sqrt{2}x = 1 \: \: or \: \: 3\sqrt{2}x = - 2$

$\rm \: x = \dfrac{1}{ \sqrt{2} } \: \: or \: \: x = - \dfrac{2}{3 \sqrt{2} }$

$\rm \: x = \dfrac{1}{ \sqrt{2} } \times \frac{ \sqrt{2} }{ \sqrt{2} } \: \: or \: \: x = - \dfrac{ \sqrt{2} . \sqrt{2} }{3 \sqrt{2} }$

$\rm \: x = \dfrac{ \sqrt{2} }{2} \: \: or \: \: x = - \dfrac{ \sqrt{2} }{3}$

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

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 = c.

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

Answer 2

Question :

Find the Roots of 6x² - √2x - 2 = 0

Solution :

by using Quadratic formula we can find roots

$\boxed{ \color{gold} \bf x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} }$

then substitutes the values.

$\rm⇢ \: \: x = \frac{ - ( - \sqrt{2}) \pm \sqrt{ {( \sqrt{2} )}^{2} - 4 \times 6 \times - 2 } }{2 \times 6}$

$\rm⇢ \: \: x = \frac{ \sqrt{2} \pm \sqrt{2 + 48} }{12}$

$\rm⇢ \: \: x = \frac{ \sqrt{2} \pm \sqrt{50} }{12}$

Final Answer :

The Roots are

$\rm \red{ x = \frac{ \sqrt{2} + \sqrt{50} }{12} } \: \: and \: \: \red{ x = \frac{ \sqrt{2} - \sqrt{50} }{12} }$

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Formula Used :

$\boxed{ \color{lime} \bf x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} }$