Question

Find the roots of the equation $2x^2-5x+3=0$, by factorization.

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

We need to find two numbers whose product is $(3\times 2)=6$ and whose sum is -5.
Let us first split the middle term $-5xas-2x-3x[\therefore (-2x)\times (-3x)=6x^2=(2x^2)\times 3]$.
So, $2x^2-5x+3=2x^2-2x-3x+3=2x(x-1)-3(x-1)=(2x-3)(x-1)$
Now, $2x^2-5x+3=0$ can be rewritten as $(2x-3)(x-1)=0$.
So, the values of x for which $2x^2-5x+3=0$ are the same for which (2x - 3)(x - 1) = 0,
i.e., either 2x - 3 = 0 or x - 1 = 0.
Now, 2x - 3 = 0 gives $x=\frac{3}{2}$
and x - 1 = 0 gives x = 1.
So, $\frac{3}{2}$ and 1 are the roots of the quadratic equation $2x^2-5x+3=0$