1. The first step is to find the discriminant itself. Now, the discriminant of a quadratic equation in the form y = ax^2 + bx + c is given by:
Ξ = b^2 - 4ac
Our equation is y = x^2 + 3x - 10. Thus, if we compare this with the general quadratic equation I outlined in the first line, we would find that a = 1, b = 3 and c = -10. It is easy to see this if we put the two equations right on top of one another:
y = ax^2 + bx + c
y = (1)x^2 + 3x - 10
Now that we know that a = 1, b = 3 and c = -10, we can substitute this into the formula for the discriminant we defined before:
Ξ = b^2 - 4ac
Ξ = (3)^2 - 4(1)(-10) (Substitute a = 1, b = 3 and c = -10)
Ξ = 9 + 40 (-4*(-10) = 40)
Ξ = 49 (Evaluate 9 + 40 = 49)
Thus, the discriminant is 49.
2. The question itself asks for the number and nature of the solutions so I will break down each of these in relation to the discriminant below, starting with how to figure out the number of solutions:
β’ There are no solutions if the discriminant is less than 0 (ie. it is negative).
If you are aware of the quadratic formula (x = (-b Β± β(b^2 - 4ac) ) / 2a), then this will make sense since we are unable to evaluate β(b^2 - 4ac) if the discriminant is negative (since we cannot take the square root of a negative number) - this would mean that the quadratic equation has no solutions.
β’ There is one solution if the discriminant equals 0.
If you are again aware of the quadratic formula then this also makes sense since if β(b^2 - 4ac) = 0, then x = -b Β± 0 / 2a = -b / 2a, which would result in only one solution for x.
β’ There are two solutions if the discriminant is more than 0 (ie. it is positive).
Again, you may apply this to the quadratic formula where if b^2 - 4ac is positive, there will be two distinct solutions for x:
-b + β(b^2 - 4ac) / 2a
-b - β(b^2 - 4ac) / 2a
Our discriminant is equal to 49; since this is more than 0, we know that we will have two solutions.
Now, given that a, b and c in y = ax^2 + bx + c are rational numbers, let us look at how to figure out the number and nature of the solutions:
β’ There are two rational solutions if the discriminant is more than 0 and is a perfect square (a perfect square is given by an integer squared, eg. 4, 9, 16, 25 are perfect squares given by 2^2, 3^2, 4^2, 5^2).
β’ There are two irrational solutions if the discriminant is more than 0 but is not a perfect square.
49 = 7^2, and is therefor a perfect square. Thus, the quadratic equation has two rational solutions (third answer).
~ To recap:
1. Finding the number of solutions.
If:
β’ Ξ < 0: no solutions
β’ Ξ = 0: one solution
β’ Ξ > 0 = two solutions
2. Finding the number and nature of solutions.
Given that a, b and c are rational numbers for y = ax^2 + bx + c, then if:
β’ Ξ < 0: no solutions
β’ Ξ = 0: one rational solution
β’ Ξ > 0 and is a perfect square: two rational solutions
β’ Ξ > 0 and is not a perfect square: two irrational solutions
