Hey guys,

I've been getting problems that have the two of the same solutions, i was wondering if this meant that there is only one solution to the quadratic equation or is there more work to be done to finish the question. ie.

1. x^2 - 6x + 9 = 0

= (x-3) (x-3)

X= 3, X= 3

example two

2. 2x^2 - 8x + 8 = 0
= 2(x^2 - 4x + 4)
= 2(x-2)(x-2)
x= 2, x=2

does this mean for both example 1 and 2 that the only solutions is 3 for #1 and 2 for number two?

thanks,

2. That's correct. The only solution to # 1 is x = 3, and the only solution for # 2 is x = 2. However, both of those solutions have multiplicity 2. They are double roots. So, to state your answer completely, e.g., you could say that x = 3 is a double root of the equation in # 1, and analogously for # 2.

3. If you were to graph these quadratics, you would see straight away that the turning point lies on the x-axis, meaning there is only one x-intercept.

Hey guys,

I've been getting problems that have the two of the same solutions, i was wondering if this meant that there is only one solution to the quadratic equation or is there more work to be done to finish the question. ie.

1. x^2 - 6x + 9 = 0

= (x-3) (x-3)

X= 3, X= 3

example two

2. 2x^2 - 8x + 8 = 0
= 2(x^2 - 4x + 4)
= 2(x-2)(x-2)
x= 2, x=2

does this mean for both example 1 and 2 that the only solutions is 3 for #1 and 2 for number two?

thanks,
By using the discriminant you can know "how many" solutions a quadratic has for

$\displaystyle ax^2+bx+c=0$

if $\displaystyle \sqrt{b^2-4ac} > 0$ there will be two distinct real solutions.

if $\displaystyle \sqrt{b^2-4ac} = 0$ there will be one real solution. This is the repeated solution that Ackbeet mentioned.

if $\displaystyle \sqrt{b^2-4ac} < 0$ there are no real solution. The solutions will be complex numbers and conjugates of each other.