According to the Fundamental Theorem of Algebra, an equation of the form a_n*x^n + a_(n-1)*x^(n-1) + ... + a_0 = 0, where a_i are coefficients and n is the degree of the equation, has exactly n solutions in the complex number system.
The equation 42x^2 + 112x - 20 = 0 can be written as 21x^2 + 56x - 10 = 0 by dividing each term by 2.
Using the quadratic formula, x = (-b ± √(b^2 - 4ac)) / 2a, where a, b, and c are the coefficients of the equation, we can find the solutions:
For x = (-56 ± √(56^2 - 4(21)(-10))) / (2*21),
x = (-56 ± √(3136 + 840)) / 42,
x = (-56 ± √(3976)) / 42,
x = (-56 ± 62.98) / 42.
The two possible values of x are:
x = (-56 + 62.98) / 42 ≈ 0.165,
x = (-56 - 62.98) / 42 ≈ -2.994.
Therefore, the equation 42x^2 + 112x - 20 = 0 has two solutions, but neither of them is an integer.
According to the Fundamental Theorem of Algebra, 42? + 112 - 20 = 0 has two solutions. How many of those solutions
are integers?
1 answer