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Find the constant k such that the quadratic 2x^2 + 3x + k has a double root

 Apr 5, 2020
 #1
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+1

try       2x2 + 3x +2       Factors to   (2x+2)(x+1)         roots = -1   and  -1

 Apr 5, 2020
 #3
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No,  (2x+2)(x+1) multiplies out to 2x2 + 4X + 2 

Guest Apr 5, 2020
 #2
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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

 Apr 5, 2020

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