x = - b ± sqrt(b^2-4ac)/2a
The quantity under the radical sign is known as the discriminant of the quadratic equation. We can identify the parameters a, b, and c for the equation x^2-6x+5=0 by comparing it to the general standard form of a quadratic equation.
ax^2+ bx+c=0
1x^2+( -6)x+5=0
As we can see, for this case a = 1, b= - 6, and c=5. Let's calculate the discriminant value.
As we can see, we have two solutions, x=1 and x=5.
b
We will now follow the same steps as in Part A. First we will compare the given equation x^2+x-20=0, to the general standard form of a quadratic equation to identify the parameters.
ax^2+ bx+c=0
1x^2+( 1)x+(- 20)=0Note that a = 1, b= 1, and c=- 20. With this information we can calculate the discriminant's value.
Let's see what happens when the discriminant is a perfect square. If n is an integer, then n^2 is a perfect square.
x = - b ± sqrt(n^2)/2a ⇔ x = - b ± n/2a
In a case like this, as long as a and b are both rational numbers the numerator - b ± n will be a rational number, since a rational number plus another rational number always gives a rational number. Recall that any integer is also rational. The denominator, 2a, is rational as well since the product of two rational numbers is rational.
x = - b ± n^(rational)/2a_(rational)
Finally, since the quotient of two rational numbers is rational, x would be a rational number under these conditions. We could see this happening in Parts A and B.
