Question

While finding the roots of the quadratic equation 3$x^2$- 2$\sqrt{6}$ x + 2 = 0 using factorization method, the given equation is reduced to the form (ax+b)(cx+d). The values of a and d are

• A. √3, √6
• B. √3, -√2
• C. -√3, √6
• D. -√7, √6

Answer 1

The correct option is B

√3, -√2

If a$x^2$+bx+c=0,

Find two numbers p,q such that p+q = b and pq = (a $\times$ c)

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

We need to split the middle term,

We can write as 3$x^2$- $\sqrt{6}$x- $\sqrt{6}$x + 2=0

Take the common terms out

$\sqrt{3}$x( $\sqrt{3}$x- $\sqrt{2}$)- $\sqrt{2}$($\sqrt{3}$x- $\sqrt{2}$ )=0

⇒( $\sqrt{3}$x- $\sqrt{2}$) ( $\sqrt{3}$x- $\sqrt{2}$ )=0

So the roots can be found out by equating the factors to zero.

Therefore the roots will be x=$\sqrt{\left(\frac{2}{3}\right)}$ , $\sqrt{\left(\frac{2}{3}\right)}$

Comparing the co-effecients of x with (ax+b)(cx+d), we get a=$\sqrt{3}$ , d=-$\sqrt{2}$