Question

Describe the nature of the roots for this equation

x^2-2x+1=0

and this one

x^2+3x-7=0

Answer 1

Hm, do you know the quadratic formula?

Answer 2

no

Answer 3

well let's begin with the first one.
\[
x^2 -2x + 1 = 0 \\
(x - 1)(x-1) = 0 \\
(x-1)^2 = 0 \;// \sqrt{}\\
x-1 = 0 \\
x = 1
\]

Answer 4

ok

Answer 5

so would it be one real double root?

Answer 6

For the second I don't think you can do this without the formula. Because 7 is a prime number, we can't simplify it that way.
The formula looks like that
\[
ax^2 + bx + c = 0 \\
x_{1,2} = \frac{ -b \pm \sqrt{b^2 - 4ac} }{ 2a }
\]
I can write down a proof if you wish, it's not complicated.

And yes, there is only one x value that makes it be equal to 0.

Answer 7

Perhaps a better explanation to what I meant up there.
Since the only 2 numbers that can multply up to 7 are 1 and 7
and writing this
\[
(x-1)(x+7) = x^2 -x +7x - 7 = x^2 +6x - 7\\
(x+1)(x-7) = x^2 + x +-7x - 7 = x^2 -6x - 7
\]
we can only construct those two...