Question

Solve it now please:– 3x² - 2√6x + 2 = 0​

Answer 1

Question:

• $\sf{3x {}^{2} - 2 \sqrt{6} + 2 = 0 }$

What we have to calculate:

• We have to calculate and find out the value of x. Here the values of x would be two as it is a quadratic equation.

Formula we are going to use:

The roots of the equation can be obtained by using this formula,

• $\red{ \boxed{ \sf{x \: = \: \dfrac{ - b± \sqrt{b {}^{2} - 4ac } }{2a} }}}$

Lets solve it:

Here as we can clearly see that b is -2√6 , a is 3 and c is 2.

Substituting the values,

$: \longmapsto \sf{x \: = \: \dfrac{ - ( - 2 \sqrt{6} ) \: ± \: \sqrt{( - 2 \sqrt{6} ) {}^{2} } - 4(3)(2) }{2(3)} }$

We know that two minus signs gives plus (addition)

$: \longmapsto \sf{x \: = \: \dfrac{ - ( - 2 \sqrt{6} ) \: ± \: \sqrt{( - 2 \sqrt{6} ) \times ( - 2 \sqrt{6}) } - 4(3)(2) }{2(3)} }$

$: \longmapsto \sf{x \: = \: \dfrac{ - ( - 2 \sqrt{6} ) \: ± \: \sqrt{( - 2 \sqrt{6} ) \times ( - 2 \sqrt{6}) } - 4 \times 3 \times 2 }{2(3)} }$

$: \longmapsto \sf{x \: = \: \dfrac{ - ( - 2 \sqrt{6} ) \: ± \: \sqrt{( - 2 \sqrt{6} ) \times ( - 2 \sqrt{6}) } - 4 \times 3 \times 2 }{2 \times 3} }$

$: \longmapsto \sf{x \: = \: \dfrac{2 \sqrt{6} ± \sqrt{24 - 12 \times 2} }{2 \times 3} }$

$: \longmapsto \sf{x \: = \: \dfrac{2 \sqrt{6} ± \sqrt{24 - 24} }{6} }$

$: \longmapsto \sf{x \: = \: \dfrac{2 \sqrt{6} }{6} }$

Cancelling it,

$: \longmapsto \sf{x \: = \: \cancel\dfrac{2 \sqrt{6} }{6} }$

Now,

$: \longmapsto \sf{x \: = \: \dfrac{2}{ \sqrt{6} } }$

Again we gets,

$: \longmapsto \sf{x \: = \: \sqrt{ \dfrac{4}{6} } }$

Cancelling,

$: \longmapsto \sf{x \: = \: \sqrt{ \cancel \dfrac {4}{6} } }$

$: \longmapsto \: \red{ \boxed{ \sf{x \: = \: \sqrt{ \dfrac{2}{3} } }}}$

• Hence, value of x is $\sf{ \sqrt{ \dfrac{2}{3} } }$. Both the values of the quadratic equation would be same.