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Question

Question: Solve \(3{x^2} - \sqrt 6 x + 2 = 0\) ....

Solve 3x26x+2=03{x^2} - \sqrt 6 x + 2 = 0 .

Explanation

Solution

Before you start solving the equation in the form ax2+bx+c=0a{x^2} + bx + c = 0 we need to determine the nature of the discriminant i.e, b24ac{b^2} - 4ac
If b24ac<0{b^2} - 4ac < 0 , then there will be no real solutions.
If b24ac>0{b^2} - 4ac > 0 , then there will be two real solutions.
If b24ac=0{b^2} - 4ac = 0 , then there will be one real solution.
If b24ac0{b^2} - 4ac \geqslant 0 then proceed further to solve for x using x=b±b24ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}

Stepwise Solution:
Given: 3x26x+2=03{x^2} - \sqrt 6 x + 2 = 0.
Comparing given eqaution with the general form of quadratic equation ax2+bx+c=0a{x^2} + bx + c = 0 we get,
From the above equation a is 33 , b is 6- \sqrt 6 and c is 22 .
Discriminant b24ac{b^2} - 4ac is as follows:
=(6)24×3×2= {\left( { - \sqrt 6 } \right)^2} - 4 \times 3 \times 2
=624 =18 18<0  = 6 - 24 \\\ = - 18 \\\ \Rightarrow - 18 < 0 \\\
As b24ac<0{b^2} - 4ac < 0, there will be no real solutions.

Note: In such types of questions which involve the concept of finding a solution to a given equation we will need to have knowledge about discriminant and formula to find the value of x. As calculations play a critical role in these kinds of questions we will need to be vigilant about that while solving.