Find the constant k such that the quadratic 2x^2 + 3x + k has a double root
+46 For this quadratic to have a double root(where there is only one solution for x), the disriminant must be 0
In the quadratic formula, the discriminant is the stuff in the square root, which is \(\sqrt{b^2-4ac}\)
To find a, b, and c, put the quadratic in the form ax^2+bx+c(it is already in this form)
So,
a=2
b=3
and c=k
Now plug the numbers in to get
\(\sqrt{3^2-4(2)(k)}\)
which equals
\(\sqrt{9-8k}\)
The stuff inside must equal zero, so the equation becomes
\(9-8k=0\)
Then solve for k:
Subtract nine on both sides
-8k=-9
k=9/8
or 1.125
XD
