To find the solutions to the equation 4x^2 + 11x - 20 = 0, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac))/(2a)
In this case, a = 4, b = 11, and c = -20. Plugging in these values, we get:
x = (-11 ± √(11^2 - 4*4*(-20)))/(2*4)
= (-11 ± √(121 + 320))/(8)
= (-11 ± √441)/(8)
= (-11 ± 21)/(8)
This gives us two possible solutions:
1) x = (-11 + 21)/8 = 10/8 = 5/4 = 1.25
2) x = (-11 - 21)/8 = -32/8 = -4
Only one of these solutions, -4, is an integer.
According to the Fundamental Theorem of Algebra, 4x2+11x−20=0
has two solutions. How many of those solutions are integers?
1 answer