Need a hint: Find k such that the equation

Question

anonymous · Answer

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 · Answer

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 · Answer

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 $D < 0$

anonymous · Answer

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

sirm3d · Answer

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

sirm3d · Answer

repeated real means the real roots are the same.

anonymous · Answer

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

asnaseer · Answer

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 · Answer

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 · Answer

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 · Answer

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 · Answer

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 · Answer

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

asnaseer · Answer

yw :)