SOLUTION: The question from my book says that "the radicand of the quadratic formula, b^2 - 4ac, can be used to determine whether {{{ ax^2 + bx + c = a}}} has solutions that are rational, ir

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: The question from my book says that "the radicand of the quadratic formula, b^2 - 4ac, can be used to determine whether {{{ ax^2 + bx + c = a}}} has solutions that are rational, ir      Log On


   

Question 541963: The question from my book says that "the radicand of the quadratic formula, b^2 - 4ac, can be used to determine whether +ax%5E2+%2B+bx+%2B+c+=+a has solutions that are rational, irrational, or not real numbers." I haven't been able to solve the problem on this website because it's saying that the first coefficient is zero. It's also asking me to explain how this works. And if it possible to determine the kinds of answers that one will obtain to a quadratic equation without actually solving the equation. I would greatly appreciate any help I can get.
Found 2 solutions by neatmath, josmiceli:
Answer by neatmath(302) About Me  (Show Source):
You can put this solution on YOUR website!

Well, if the first coefficient is zero, then you would not have a quadratic equation, but a linear equation, ie bx%2Bc=a

Now, there are specific rules for using the sqrt%28b%5E2-4ac%29

a.) If sqrt%28b%5E2-4ac%29%3E0 you will have 2 real solutions.

b.) If sqrt%28b%5E2-4ac%29=0 you will have 1 real solution.

c.) If sqrt%28b%5E2-4ac%29%3C0 you will have 2 imaginary solutions.

Those are the only possible scenarios you can have with a quadratic equation.

In part c, you will end up with a negative number under the radical, thus the reason for having only imaginary solutions.

You would not know whether your solutions in part a would be rational or irrational until you simplified the radical.

There is a lot more to using the quadratic formula, but I hope that this is a good start.

There are many resources available online for learning more about the quadratic formula.

I hope this helps!

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The quadratic formula is:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
I will make the replacement
+sqrt%28b%5E2++-+4a%2Ac%29+=+z+
Now I have
+x+=+%28-b+%2B+z%29+%2F+%282%2Aa%29+
and
+x+=+%28-b+-+z%29+%2F+%282%2Aa%29+
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If z is not zero and is a real number, I
get a different solution for each equation,
so 2 solutions.
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If z+=+0, then there is only 1 solution
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Also, since z is a square root, it can be
+sqrt%28-1%2An%29+, which is +i%2Asqrt%28n%29+
and there are 2 imaginary solutions.
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These possibilities depend on what
+b%5E2+-+4a%2Ac+ is
+b%5E2+-+4a%2Ac+ = 0 ( one solution )
+b%5E2+-+4a%2Ac+ = (+) ( two real solutions )
+b%5E2+-+4a%2Ac+ = (-) ( two imaginary solutions )
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In the case of +b%5E2+-+4a%2Ac+ = (+), if this is a perfect square,
like +64+ The 2 real solutions will be rational, which means
a fraction of some kind.
If it is not a perfect square, like 31, the 2 real solutions
will be irrational, like +3.666666+. . . forever
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Hope this isn't confusing.