"One rational solution" $._("One rational solution") randRangeNonZero(-6, 6) * 2 randFromArray(getFactors(B * B / 4)) (B * B) / (4 * A) B * B - 4 * A * C

Describe the solutions to the following quadratic equation:

expr(["+", ["*", A, ["^", "x", 2]], ["*", B, "x"], C]) = 0
ANSWER
  • One rational solution
  • Two rational solutions
  • Two irrational solutions

We could use the quadratic formula to solve for the solutions and see what they are, but there's a shortcut...

\qquad x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

KhanUtil.processMath($("#quadratic")[0], "\\qquad x = \\dfrac{-b \\pm" + " \\sqrt{\\blue{b^2 - 4ac}}}{2a}", true);
Think about what the part of the quadratic formula under the radical tells us about the solutions.

Substitute the a, b, and c coefficients from the quadratic equation:

\qquad\begin{array} && b^2-4ac \\ \\ =& (B)^2 - 4 ( A)(C) \\ \\ =& DISCRIMINANT \end{array}

Because \blue{b^2 - 4ac} = 0, then the quadratic formula reduces to \dfrac{-b}{2a}, which means there is just one rational solution.

Because \blue{b^2 - 4ac} is a perfect square, its square root is rational and the quadratic formula reduces to \dfrac{-b \pm sqrt(DISCRIMINANT)}{2a} , which means there are two rational solutions.

Because \blue{b^2 - 4ac} is not a perfect square, its square root is irrational and the quadratic formula reduces to \dfrac{-b \pm \sqrt{DISCRIMINANT}}{2a} , which means there are two irrational solutions.

"Two rational solutions" $._("Two rational solutions")
randRangeNonZero(-9, 9) randRangeNonZero(-9, 9) randRangeNonZero(-9, 9) B * B - 4 * A * C
"Two irrational solutions" $._("Two irrational solutions")
randRangeNonZero(-9, 9) randRangeNonZero(-9, 9) randRangeNonZero(-9, 9) B * B - 4 * A * C