with the polynomial: 2x^2 + 7x - 15 = 0
do these apply?
There are 2 real roots.
There are 2 irrational roots.
well, both can't be true, can they ?
Hint, have you calculated the discriminant ?
Yes it is 169
Oh, so there are 2 real roots?
You got it, and to top it off, since 169 is a perfect square they are even rational
(still real though)
To determine the number and types of roots for the given polynomial, you can use the discriminant. The discriminant is a value calculated using the coefficients of the polynomial, and it helps determine the nature of the roots.
The discriminant (denoted as Δ) is equal to b^2 - 4ac, where a, b, and c are the coefficients of the quadratic polynomial (ax^2 + bx + c = 0).
In the given polynomial, the coefficients are:
a = 2
b = 7
c = -15
Now, let's calculate the discriminant:
Δ = (7)^2 - 4(2)(-15)
= 49 + 120
= 169
The value of the discriminant is Δ = 169.
Based on the value of the discriminant:
1. If Δ > 0, there are two real and distinct roots.
2. If Δ = 0, there is only one real root (a repeated root).
3. If Δ < 0, there are two complex roots (conjugate pairs) with no real part.
In this case, Δ = 169, which is greater than zero (Δ > 0). Therefore, the given polynomial has two distinct real roots.
However, for the given polynomial, it's not possible to determine whether there are two irrational roots since the discriminant only provides information about the type of roots (real, repeated real, or complex), not the exact values of the roots. To find the exact values of the roots, you can use the quadratic formula:
x = (-b ± √Δ) / (2a)
Plugging in the coefficients of the polynomial (a = 2, b = 7, and c = -15) into the quadratic formula will give you the exact values of the roots.