1. consider the following quadratic equation x^2 =4x -5. how many solutions does it have?
a the equation has one real solution.
b the equation has two real solutions.
c the equation has no real solutions
d the number of solutions can not be determined.

2. consider the quadratic equation ax^2+bx+c where a, b, and c are rational numbers and the quadratic has two distinct zeros. if one is rational, what is true for the other zeros.

a) the other zero is rational.
b) the other zero could be rational or irrational.
c) the other zero is not a real number.
d) the other zero is rational.

         D) The other zero is rational.

Step-by-step explanation:

1. The attached graph shows there are no values of x where the left side of the equation is equal to the right side of the equation. There are no real solutions.

If you rewrite the quadratic to standard form (by subtracting the right-side expression), you get ...

  x^2 -4x +5 = 0

Using the hint, the discriminant is b^2-4ac = (-4)^2-4(1)(5) = 16 -20 = -4. It is negative, so both roots will be complex.

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2. The (rational) number "b" is the opposite of the sum of the zeros. If one zero is rational, the other root must be. The difference of rational numbers is rational.

Choices A and D are identical, so both are correct. Perhaps a typo?

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We can also consider the discriminant in problem 2. The roots of the quadratic are the sum (or difference) of -b and the square root of the discriminant, all divided by 2a. Any one root can only be rational if the square root is rational. In that case, the other root will be rational as well.

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