Lesson goal: Using the Quadratic Formula

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Remember solving quadractic equations? If you have a quadractic equation in "standard form," which is $Ax^2+Bx+C=0$, then the two solutions to this equation will be $$x=\frac{-B\pm\sqrt{B^2-4AC}}{2A}.$$

In grade-school algebra the equations are kept simple in that the roots (or answers) to quadractic equations are always real. If you look carefully at the quadractic formula, there is a $\sqrt{B^2-4AC}$. As you know, you cannot take the square root of a negative number (if you want roots that are real numbers). In other words, $B^2-4AC$ (which is called the discriminant) must always be $\ge 0$, so in a quadratic equation solver, we must check that this is true (with an if statement) before continuing.

Now you try. Put some inequality to test in the if statement. You also have to complete the x1= and x2= lines to give the $+$ and $-$ roots, as computed with the quadratic formula above.

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This code will not run! What will you put in the if ???? then statement to ensure that $d$ (the discriminant) is $\ge 0$ before continuing to compute $x1$ and $x2$, the two solutions to the equation?

By the way, in this case we'll say that $x1=\frac{-B+\sqrt{B^2-4AC}}{2A}$ and $x2=\frac{-B-\sqrt{B^2-4AC}}{2A}.$ Dismiss.

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