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Need a hint: Find k such that the equation

Mathematics
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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.
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'.
the quadratic formula should also work. in \[x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\] 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 \(D < 0\)

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Other answers:

Maybe I'm having trouble understanding terms. What do they mean when they say repeated real?
so go ahead use the quadratic formula, the set the expression under the radical sign (equal) to zero, and solve k.
repeated real means the real roots are the same.
Ah, that helps, sirm3d. Ok, I'm going tot take a crack at it.
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:\[k(x-r_1)(x-r_1)=0\]
Serious brain farts today...I'm having trouble seeing how you factored a generic term like \[x^2+\frac{1}{k}x+1=0\]
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:\[(x+a)^2\]
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. :)
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.
Ah, I'm going to take note of that. Thanks again, both.
yw :)

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