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eighthourlunch Group TitleBest ResponseYou've already chosen the best response.0
Find k such that the equation \[kx^2+x+k=0\]has a repeated real solution. I thought I'd take a stab at it with the quadratic equation, tried solving for k, but I'm not sure how the book got to +/ 1/2.
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
if factor out the k you can write this as:\[k(x^2+\frac{x}{k}+1)=0\]then, since it is a repeated root, you know the terms in the braces can be written as:\[(x+a)^2\]where 'a' is some constant. Expand this out and compare like terms to work out the value for 'k'.
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.1
the quadratic formula should also work. in \[x=\frac{ b \pm \sqrt{b^24ac} }{ 2a }\] the discriminant \(D=b^2  4ac\) says i. 2 distinct roots if \(D>0\) ii. 2 equal (repeated) roots if \(D = 0\) iii. 2 complex (conjugate) roots if \(D < 0\)
 one year ago

eighthourlunch Group TitleBest ResponseYou've already chosen the best response.0
Maybe I'm having trouble understanding terms. What do they mean when they say repeated real?
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.1
so go ahead use the quadratic formula, the set the expression under the radical sign (equal) to zero, and solve k.
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.1
repeated real means the real roots are the same.
 one year ago

eighthourlunch Group TitleBest ResponseYou've already chosen the best response.0
Ah, that helps, sirm3d. Ok, I'm going tot take a crack at it.
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
repeated real means this has two real solutions that are identical. so lets say the roots of this equation are both \(r_1\), then we can write it as:\[k(xr_1)(xr_1)=0\]
 one year ago

eighthourlunch Group TitleBest ResponseYou've already chosen the best response.0
Serious brain farts today...I'm having trouble seeing how you factored a generic term like \[x^2+\frac{1}{k}x+1=0\]
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
I didn't factor it  instead I used the information given in the question "it has repeated real roots" to infer that it must therefore be of the form:\[(x+a)^2\]
 one year ago

eighthourlunch Group TitleBest ResponseYou've already chosen the best response.0
So from there, knowing they're repeated, you say a, which is 1/k at this point would have to be 1/2 or 1/2, because that's the only answer that will give you \[x^2+2x+1\] Ok, that makes sense. Thank you. I think I've been getting to cocky working out physics and calculus. Good thing this is just review. :)
 one year ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.1
although the method proposed by @sirm3d looks like a simpler approach  i.e. make use of the fact that the discriminant D=0 for repeated roots.
 one year ago

eighthourlunch Group TitleBest ResponseYou've already chosen the best response.0
Ah, I'm going to take note of that. Thanks again, both.
 one year ago
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