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A quadratic is any equation whose highest power is \(x^2\). Every one graphs as a parabola, and every one can be solved — factoring is fastest when it works, and the quadratic formula never fails.
Get everything on one side equal to zero first — every method below assumes that.
If it factors, set each factor to zero (zero-product property): \(x = 2\) or \(x = 3\). Fast, but not everything factors nicely.
When there is no plain \(x\) term, isolate the square and take the root of both sides. Do not forget the \(\pm\).
Rewrite as \((x + p)^2 = q\), then square-root. Always works, and it is where the quadratic formula comes from.
Plug \(a\), \(b\), \(c\) straight in. This always works — reach for it whenever factoring is not obvious.
Opens up when \(a > 0\), down when \(a < 0\). The vertex sits on the axis of symmetry at \(x = -\tfrac{b}{2a}\); the real solutions are where the parabola crosses the x-axis.
Solve: \(x^2 - 5x + 6 = 0\)
Two numbers that multiply to \(+6\) and add to \(-5\): \(-2\) and \(-3\).
Set each factor to zero.
Solve: \(2x^2 - 50 = 0\)
Isolate the square term.
Take the square root of both sides — keep the \(\pm\).
Solve: \(x^2 + 4x + 1 = 0\)
Identify \(a = 1\), \(b = 4\), \(c = 1\) and plug into the formula.
Simplify under the root: \(16 - 4 = 12\), and \(\sqrt{12} = 2\sqrt{3}\).
Divide every term by 2.
Solve each equation, then reveal the solution to check yourself.
\(x^2 - 9 = 0\)
Square-root method: \(x^2 = 9\).
\(x^2 + 7x + 10 = 0\)
Factors: \(+2\) and \(+5\) multiply to 10 and add to 7.
\(x^2 - 6x + 9 = 0\)
Perfect-square trinomial; the discriminant is 0, so one repeated root.
\(3x^2 - 12 = 0\)
Isolate: \(3x^2 = 12 \Rightarrow x^2 = 4\).
\(x^2 + 2x - 4 = 0\)
Does not factor nicely — use the formula with \(a=1, b=2, c=-4\). Discriminant \(= 4 + 16 = 20\), \(\sqrt{20} = 2\sqrt{5}\).
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