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- anonymous

Need a hint: Find k such that the equation

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- anonymous

Need a hint: Find k such that the equation

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- anonymous

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.

- asnaseer

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'.

- sirm3d

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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- anonymous

Maybe I'm having trouble understanding terms. What do they mean when they say repeated real?

- sirm3d

so go ahead use the quadratic formula, the set the expression under the radical sign (equal) to zero, and solve k.

- sirm3d

repeated real means the real roots are the same.

- anonymous

Ah, that helps, sirm3d. Ok, I'm going tot take a crack at it.

- asnaseer

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:\[k(x-r_1)(x-r_1)=0\]

- anonymous

Serious brain farts today...I'm having trouble seeing how you factored a generic term like \[x^2+\frac{1}{k}x+1=0\]

- asnaseer

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:\[(x+a)^2\]

- anonymous

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. :)

- asnaseer

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.

- anonymous

Ah, I'm going to take note of that. Thanks again, both.

- asnaseer

yw :)

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