Thread: A quadratic equation and its real roots

The quadratic equation kx^2+(k-3)x+1=0 has two equal real roots.
a) Find the possible values of k.
b) Write down the values of k for which x^2+(k-3)x+k=0 has two equal real roots.

I solved a) using the quadratic formula and found k=1 or k=9, but I don't know how to do b). If I use the quadratic formula, then the values for k are the same as in a). Is that correct? Does k=1 or 9 for b) as well?

Originally Posted by bhuang

The quadratic equation kx^2+(k-3)x+1=0 has two equal real roots.
a) Find the possible values of k.
b) Write down the values of k for which x^2+(k-3)x+k=0 has two equal real roots.

I solved a) using the quadratic formula and found k=1 or k=9, but I don't know how to do b). If I use the quadratic formula, then the values for k are the same as in a). Is that correct? Does k=1 or 9 for b) as well?

Yes the answers for (a) and (b) are the same.

When you found k = 1 OR k = 9, you got there because you concluded B^2 - 4AC = 0, right? So since both quadratic equations have the same B^2 - 4AC, the solutions are the same.

Originally Posted by bhuang

b) Write down the values of k for which x^2+(k-3)x+k=0 has two equal real roots.

I think they want you to find where