# Quadratic Problem- 25 Dec

**Question**

If !!Sinalpha!! is a root of !!25x^2 + 5x - 12 = 0!!, !!-1 < x < 0!!

Then !!Cos2alpha = ?!!

!!a)" "0!!

!!b)" "7/25!!

!!c)-7/25!!

!!d)" "!!None

**Solution**

Given equation !!25x^2 + 5x - 12 = 0!!

!!=> 25x^2 + 20x - 15x - 12 = 0!!

!!=> (5x - 3)(5x + 4) = 0!!

This will give

!!x = 3/5!! and !!-4/5!!

Since given condition !!-1 < x < 0!!

Hence, taking negative value, !!x = Sinalpha = -4/5!!

Now we know that, !!Cos2alpha = 1 - 2Sin\^2 alpha!!

So, !!Cos2alpha = 1 - 2(-4/5)^2!!

!!=> Cos2alpha = 1 - 2(16/25) = (25-32)/25 = -7/25!!