SOLUTION: Use the discriminant to determine whether each quadratic equation has two unequal real solutions,a repeated real solution,or no real solution,without solving the equation. x

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Question 1207406: Use the discriminant to determine whether each quadratic equation has two unequal real solutions,a repeated real solution,or no real solution,without solving the equation.


x^2 + 4x + 7 = 0


I know the discriminant is the radicand of the quadratic formula.


So, I need to use b^2 - 4ac where a = 1, b = 4 and c = 7.


b^2 - 4ac


(4)^2 - 4(1)(7)

16 - 28 = -12


The answer is -12.


The discriminant is negative, which means the equation has no real solution.


You say?


What would be an example of a repeated real solution or no solution at all?

Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52775) About Me  (Show Source):
You can put this solution on YOUR website!
.

(1)  Yes.



(2)  Example of repeated solutions

        %28x-1%29%5E2 = 0,

     which is the same as

        x%5E2+-+2x+%2B+1 = 0.



(3)  Example of "no solutions"

        %28x-1%29%5E2 + 1 = 0,

     which is the same as

        x%5E2+-+2x+%2B+2 = 0.

Solved, answered and explained.



Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


Note it is possible, or even probable, that you have not asked the second question correctly.

As implied in the statement of the problem, every quadratic equation has two unequal real solutions, a repeated real solution, or no real solutions. There is no quadratic equation that has "no solution at all".

The given equation has no real solutions: b^2-4ac = 16-28 = -12 < 0.

By definition, any perfect square quadratic polynomial has a repeated real solution (or, better stated, two equal solutions): x^2-6x+9 --> b^2-4ac = 36-36 = 0.

For an example with two real solutions, start with any quadratic polynomial that is a factor of two linear polynomials. Example: (x+3)(x-2) = x^2+x-6 --> b^2-4ac = 1+24 = 25 > 0.