Calculate the discriminant to determine the number solutions. y = x ^2 + 3x - 10 two irrational solutions not solutions two rational solutions one rational solution

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:

Δ = 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:

Δ = b^2 - 4ac

Δ = (3)^2 - 4(1)(-10) (Substitute a = 1, b = 3 and c = -10)

Δ = 9 + 40 (-4*(-10) = 40)

Δ = 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:

• 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 ± √(b^2 - 4ac) ) / 2a), then this will make sense since we are unable to evaluate √(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.

• 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 √(b^2 - 4ac) = 0, then x = -b ± 0 / 2a = -b / 2a, which would result in only one solution for x.

• 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 + √(b^2 - 4ac) / 2a

-b - √(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:

• 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).

• 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:

• Δ < 0: no solutions

• Δ = 0: one solution

• Δ > 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:

• Δ < 0: no solutions

• Δ = 0: one rational solution

• Δ > 0 and is a perfect square: two rational solutions

• Δ > 0 and is not a perfect square: two irrational solutions

21/58.

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b - exponent

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