anonymous
  • anonymous
Need a hint: Find k such that the equation
Mathematics
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • 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
  • 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
  • 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 \(D < 0\)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Maybe I'm having trouble understanding terms. What do they mean when they say repeated real?
sirm3d
  • sirm3d
so go ahead use the quadratic formula, the set the expression under the radical sign (equal) to zero, and solve k.
sirm3d
  • sirm3d
repeated real means the real roots are the same.
anonymous
  • anonymous
Ah, that helps, sirm3d. Ok, I'm going tot take a crack at it.
asnaseer
  • 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
  • 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
  • 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
  • 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
  • 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
  • anonymous
Ah, I'm going to take note of that. Thanks again, both.
asnaseer
  • asnaseer
yw :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.