Question

In the method to find 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. - $\sqrt{3}$, $\sqrt{6}$
• B. $\sqrt{3}$, -$\sqrt{2}$
• C. - $\sqrt{7}$, $\sqrt{6}$
• D. $\sqrt{3}$, $\sqrt{6}$

Answer 1

The correct option is B

$\sqrt{3}$, -$\sqrt{2}$

In a$x^2$+bx+c=0, find 2 numbers p,q such that p+q = b and pq=(a $\times$ c)

Solution: 3$x^2$- 2 +2= 0

We need to split the middle term, So

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, we find that a=$\sqrt{3}$ , d=-$\sqrt{2}$