Question 118816: What are the values o k which will give 4x^2+(k+7)x+(k+4)=0 equal,real roots? Please help me?
Answer by ilana(307)   (Show Source):
You can put this solution on YOUR website! To get two equal, real roots, we need a perfect square trinomial, or an equation that can be factored into something of the form (ax+b)^2=0, as opposed to (ax+b)(cx+d)=0. As you might have learned, (ax+b)^2=0 expands to (a^2)(x^2)+2ab+b^2=0. So the equation you were given, 4x^2+(k+7)x+(k+4)=0, will change to (2x+___)^2=0.
The tricky part is dealing with all of these k terms. We have 4x^2+(k+7)x+(k+4)=0 and we want (a^2)(x^2)+2ab+b^2=0. That means b^2=k+4, so b must be sqrt(k+4). a^2=4, so a=2. And finally, 2ab=k+7, for the middle term.
So:
a=2
b=sqrt(k+4)
2ab=k+7
So, using the first 2 equations in the third, 2(2)(sqrt(k+4))=k+7
4sqrt(k+4)=k+7
Square all of these terms to get rid of the sqrt
16(k+4)=(k+7)^2
16k+64=k^2+14k+49
Get all of the terms to the right side (because I want k^2 to stay k^2, and not to become -k^2)
0=k^2-2k-15
Factor the expression on the right
0=(k-5)(k+3)
k=5,-3
You can test these values in your original equation, and you should get two perfect square trinomials. Good luck!
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