Thread: Find the value of equation with equal roots

1. Find the value of equation with equal roots

I'm trying to solve this equation, but have no clue how I'm supposed to do it. My teacher tends to give out homework much more difficult than we've done in class.

Find the value of k which results in the equation having equal roots, given that k <> 0
$kx^2 + 2kx - 1 = 0$

2. Use the fact that if the equation in the form $ax^2 + bx + c$ has equal roots, then you have:

$b^2 - 4ac = 0$

Sub in and solve for k.

3. Thank you Unknown!

4. You're welcome

5. Yes. Here is another way. The polynomial $x^2+2x-1/k$ has roots iff it can be factored as $(x-a_1)(x-a_2)$ for some $a_1, a_2$. Here $a_1, a_2$ are, in fact, the roots. The right-to-left direction is obvious. One way to show the left-to-right direction is to apply the little Bézout's theorem. So if we are looking for the situation when both roots are $a$, then $(x-a)^2=x^2-2ax+a^2=x^2+2x-1/k$. Now, polynomials are equal iff their corresponding coefficients are equal, so $-2a=2$ and $a^2=-1/k$. From here, one can find $k$.

A similar way is to use Viète's formulas, which says that if $a_1, a_2$ are roots of $x^2+2x-1/k$, then $a_1+a_2=-2$ and $a_1a_2=-1/k$.

6. And, just to stick in my oar, although this is serious overkill, a function, f(x), has a double zero at x= a if and only if both f(a)= 0 and f'(a)= 0. Solve the two equations $kx^2+ 2kx- 1= 0$ and $2kx+ 2k= 0$.