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Solve for $x$: $\sqrt 3 {x^2} - 2\sqrt 2 x - 2\sqrt 3 = 0$

seo-qna
Last updated date: 15th Jul 2024
Total views: 453k
Views today: 9.53k
Answer
VerifiedVerified
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Hint- Here, we will be using a discriminant method to solve the given quadratic equation.

Given, equation is $\sqrt 3 {x^2} - 2\sqrt 2 x - 2\sqrt 3 = 0$
As we know that for any general quadratic equation $a{x^2} + bx + c = 0$, the solution is given as
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ where $d = \sqrt {{b^2} - 4ac} $ is the discriminant of the quadratic equation.
On comparing the given quadratic equation with the general quadratic equation, we get
$a = \sqrt 3 $ ,$b = - 2\sqrt 2 $ and $c = - 2\sqrt 3 $
Now substitute these values in the formula, we get
$
  x = \dfrac{{ - \left( { - 2\sqrt 2 } \right) \pm \sqrt {{{\left( { - 2\sqrt 2 } \right)}^2} - 4\left( {\sqrt 3 } \right)\left( { - 2\sqrt 3 } \right)} }}{{2\left( {\sqrt 3 } \right)}} = \dfrac{{2\sqrt 2 \pm \sqrt {8 + 24} }}{{2\sqrt 3 }} = \dfrac{{2\sqrt 2 \pm \sqrt {32} }}{{2\sqrt 3 }} \\
   \Rightarrow x = = \dfrac{{2\sqrt 2 \pm 4\sqrt 2 }}{{2\sqrt 3 }} = \dfrac{{\sqrt 2 \pm 2\sqrt 2 }}{{\sqrt 3 }} \\
 $
$ \Rightarrow {x_1} = \dfrac{{\sqrt 2 + 2\sqrt 2 }}{{\sqrt 3 }} = \dfrac{{3\sqrt 2 }}{{\sqrt 3 }} = \left( {\sqrt 3 } \right)\left( {\sqrt 2 } \right) = \sqrt 6 $ and $ \Rightarrow {x_2} = \dfrac{{\sqrt 2 - 2\sqrt 2 }}{{\sqrt 3 }} = \dfrac{{ - \sqrt 2 }}{{\sqrt 3 }} = - \sqrt {\dfrac{2}{3}} $ .
i.e., The two roots of the given quadratic equation are ${x_1} = \sqrt 6 $ and ${x_2} = - \sqrt {\dfrac{2}{3}} $.
Therefore, the two values of $x$ possible in order to satisfy the given quadratic equations are $\sqrt 6 $ and $ - \sqrt {\dfrac{2}{3}} $.

Note- For any quadratic equation, $a{x^2} + bx + c = 0$, according to the value of $d = \sqrt {{b^2} - 4ac} $ we have three possible cases:
i. If it is positive, then the quadratic equation will have two different real roots.
ii. If it is equal to zero, then the quadratic equation will have real and equal roots.
iii. If it is negative, then the quadratic equation will have two different imaginary roots.