How to we find the number of solutions in a problem like x^2-3x+1=0?

Question

I've seen many people say if its equal to 0 then it has one solution, if it is less than zero it has 2 imaginary solutions and if it is more that zero it has 2 real solutions, but how to we get to that point?

Answer 1

What you are describing sounds like using the discriminant. This is b²-4ac

1. If b²-4ac > 0 = 2 real solutions
2. If b²-4ac = 0 = 1 real solution
3. If b²-4ac < 0 = 0 real solutions (2 imaginary solutions)

For you problem,

a=1

b=-3

c=1

Plug these in

(-3)²-4(1)(1)

9-4 = 5

5>0 There are 2 real solutions.

The number of solutions to a quadratic equations is also equal to the number of x intercepts. If you graph the solution, that is the easiest way to see it.

1. 2 x-intercepts = 2 real solutions
2. 1 x-intercept = 1 real solution
3. 0 x intercepts = 0 real solutions (2 imaginary solutions)

Answer 2

Hi Jardet,

The key quantity for determining the number of solutions to a quadratic equation is the discriminant,

Δ = b² − 4ac. Recall that by the quadratic formula, the two solutions of the quadratic are: x = (−b ± √Δ) ÷ (2a). In the example you gave, a = 1, b = −3, and c = 1. Because of that square root, we end up having the three different cases.

1. Δ > 0. In this case, we're taking the square root of a positive real number, which always exists. Since there's a "±" in the formula, this will give us two real roots: x = (−b ± √Δ) ÷ (2a).
2. Δ = 0. In this case, we take the square root of zero, which is just zero. Since −b ± 0 = −b, we will only have one real root: x = −b/2a. Remember by the fundamental theorem of algebra (FTA), there are always two roots! So where's the other one? It's a repeated root: they are both equal and we can factor the polynomial as d(x + b/2a)² = 0, where 'd' is just some real number which doesn't affect the roots (you can find it by multiplying out if you care to).
3. Δ < 0. In this last case, we have to take the square root of a negative number. Since, under the standard interpretation of "square root", negative numbers don't have square roots, we introduce i² = −1 to help us solve the equation (remember, FTA says there are always two roots!). In this case, then, we will have two complex number roots: x = (−b ± i√−Δ) ÷ (2a). If it turns out that b = 0, then the two roots will be "pure" imaginary numbers.

So that's it. There are always two roots, but depending on the value of the discriminant (and hence the square root), they may be equal to two real numbers, one real number or two complex numbers!

Hope this helps,

(:

-Ben T.

Answer 3

use the quadratic formula or complete the square

x = 3/2 + or - (1/2)sqr(9-4)

= 3/2 + or - (1/2)sqr5 = about 1.5+ or - 1.1 = 2.6 or 0.4

It has 2 real solutions

if the discriminate, b^2-4ac = 0 then it has only one solution, b=-3, a=1, c=1

if the discriminate is positive then it has 2 solutions

if the discriminate is negative it has no real solutions

the discriminate = 9-4=5>0