Until now, every equation you've solved drew a straight line. A quadratic is where mathematics first bends — and the curve it traces, the parabola, hides the answer to the whole equation in plain sight.
The first curve
A quadratic equation is any equation whose highest power of \( x \) is two. Written in its standard form, it looks like this:
\[ ax^2 + bx + c = 0, \qquad a \neq 0 \]The three numbers \( a \), \( b \) and \( c \) are the coefficients. We insist that \( a \neq 0 \), because if \( a \) were zero the \( x^2 \) term would vanish and we'd be left with a plain straight-line equation — no curve at all.
Here is the secret that makes everything else fall into place. If you graph the left-hand side as a function,
\[ y = ax^2 + bx + c, \]you don't get a line — you get a smooth, symmetric curve called a parabola. When \( a > 0 \) it opens upward like a valley, a gentle \( \cup \); when \( a < 0 \) it opens downward like a hill, a \( \cap \). And the solutions of \( ax^2 + bx + c = 0 \) — the values of \( x \) we're hunting for, also called the roots — are exactly the places where that curve crosses the \( x \)-axis. Solving the equation and finding where the parabola meets the axis are the same act.
Below is a live parabola. Drag the sliders for \( a \), \( b \) and \( c \) and watch the curve respond: \( a \) controls how wide or narrow it is (and, if you make it negative, flips the whole thing over), \( c \) slides the curve straight up or down, and the dashed vertical line is the axis of symmetry running through the lowest (or highest) point, the vertex. Notice where the curve meets the \( x \)-axis — those crossing points are the roots.
Where is the turning point? The vertex of every parabola sits on its axis of symmetry, at \( x = -\dfrac{b}{2a} \). For the starting curve \( y = x^2 - 2x - 3 \) that's \( x = -\dfrac{-2}{2} = 1 \) — drag the sliders back to \( a=1,\,b=-2,\,c=-3 \) and you'll see the dashed line land exactly there, midway between the two roots.
Three ways to find the roots
Finding where the parabola crosses the axis means solving \( ax^2 + bx + c = 0 \). There are three classic methods, and it's worth knowing them in order of friendliness — reach for the easiest one that works.
1. Factoring — the quick win
Many quadratics can be rewritten as a product of two simple brackets. If you can spot the factoring, the rest is almost free, thanks to one powerful idea: if a product equals zero, at least one of its factors must be zero. So once you have
\[ (x - r)(x - s) = 0, \]the equation splits cleanly into two tiny ones, \( x - r = 0 \) and \( x - s = 0 \), giving \( x = r \) or \( x = s \). To factor \( x^2 + bx + c \), look for two numbers that multiply to \( c \) and add to \( b \). That's the whole search.
Tip Factoring only works neatly when the numbers cooperate. If you can't find an integer pair that multiplies to \( c \) and adds to \( b \) within a few seconds, don't force it — skip straight to the quadratic formula, which never fails.
2. Completing the square — the geometric trick
When factoring won't bite, you can always manufacture a perfect square. The goal is to rewrite the equation in the form
\[ (x + p)^2 = q, \]because once it looks like that, you simply take the square root of both sides and you're done. Starting from \( x^2 + bx + c = 0 \), move \( c \) across and add the magic number \( \left(\tfrac{b}{2}\right)^2 \) to both sides — exactly the amount needed to turn the left side into a perfect square:
\[ x^2 + bx + \left(\tfrac{b}{2}\right)^2 = \left(\tfrac{b}{2}\right)^2 - c, \qquad \left(x + \tfrac{b}{2}\right)^2 = \left(\tfrac{b}{2}\right)^2 - c. \]The name is literal: you are completing a square. Picture a square of side \( x \) with two thin rectangular strips of width \( \tfrac{b}{2} \) laid along its edges; the little corner square of area \( \left(\tfrac{b}{2}\right)^2 \) is the missing piece that completes the figure. This method also quietly proves the formula we're about to meet.
3. The quadratic formula — the one that always works
Complete the square on the general equation \( ax^2 + bx + c = 0 \) and out drops a formula that solves every quadratic, no cleverness required:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]The \( \pm \) is doing real work — it hands you both roots at once, the \( + \) giving one crossing point and the \( - \) the other. Memorise this one. When factoring fails and you don't feel like completing the square by hand, the formula is your reliable workhorse.
The discriminant: counting roots before you solve
Look again at the formula. Everything interesting hangs on the quantity under the square root sign. We give it its own name, the discriminant:
\[ D = b^2 - 4ac. \]Because you can't take the (real) square root of a negative number, the sign of \( D \) tells you the whole story — how many times the parabola meets the \( x \)-axis — before you do any solving at all:
- \( D > 0 \): the square root is a real positive number, so \( \pm\sqrt{D} \) gives two different answers. Two real roots — the curve crosses the axis twice.
- \( D = 0 \): the square root is zero, so the \( + \) and \( - \) collapse into the same value. One repeated root — the curve just touches the axis at its vertex.
- \( D < 0 \): there's no real square root of a negative number. No real roots — the curve floats entirely above or below the axis and never reaches it.
Go back to the widget and watch this happen. Slide \( c \) upward so the whole \( \cup \) lifts off the axis: the two crossing points rush together, briefly kiss at one point, then separate from the axis entirely. At that moment of lift-off the discriminant has gone from positive, through zero, to negative — the algebra and the picture are telling you the exact same thing.
Read before you solve. Computing \( D = b^2 - 4ac \) takes seconds and saves you from chasing roots that aren't there. A negative discriminant is not a mistake — it's the equation honestly telling you the parabola never reaches the \( x \)-axis.
- Here \( b = -5 \) and \( c = 6 \). We need two numbers that multiply to \( 6 \) and add to \( -5 \). The pair \( -2 \) and \( -3 \) works: \( (-2)(-3) = 6 \) and \( -2 + (-3) = -5 \).
- So the equation factors as \( (x - 2)(x - 3) = 0 \).
- A product is zero only when a factor is zero, so \( x - 2 = 0 \) or \( x - 3 = 0 \), giving \( x = 2 \) or \( x = 3 \).
- Verify both. At \( x = 2 \): \( 4 - 10 + 6 = 0 \). ✓ At \( x = 3 \): \( 9 - 15 + 6 = 0 \). ✓
The roots are \( x = \mathbf{2} \) and \( x = \mathbf{3} \) — the two places where the parabola \( y = x^2 - 5x + 6 \) crosses the \( x \)-axis.
- Read off the coefficients: \( a = 2 \), \( b = 3 \), \( c = -2 \).
- Compute the discriminant first: \( D = b^2 - 4ac = 3^2 - 4(2)(-2) = 9 + 16 = 25 \). It's positive, so expect two real roots.
- Substitute into the formula: \( x = \dfrac{-3 \pm \sqrt{25}}{2 \cdot 2} = \dfrac{-3 \pm 5}{4} \).
- Split the \( \pm \): \( x = \dfrac{-3 + 5}{4} = \dfrac{2}{4} = \dfrac{1}{2} \), or \( x = \dfrac{-3 - 5}{4} = \dfrac{-8}{4} = -2 \).
The roots are \( x = \mathbf{\tfrac{1}{2}} \) and \( x = \mathbf{-2} \). A quick check at \( x = -2 \): \( 2(4) + 3(-2) - 2 = 8 - 6 - 2 = 0 \). ✓
Practice
Try each one yourself, then reveal the full solution.
1. Solve \( x^2 - 7x + 12 = 0 \).
This factors nicely. We need two numbers that multiply to \( 12 \) and add to \( -7 \): the pair \( -3 \) and \( -4 \) fits, since \( (-3)(-4) = 12 \) and \( -3 + (-4) = -7 \).
So \( (x - 3)(x - 4) = 0 \), which means \( x - 3 = 0 \) or \( x - 4 = 0 \).
The roots are \( x = \mathbf{3} \) and \( x = \mathbf{4} \). Check: at \( x = 3 \), \( 9 - 21 + 12 = 0 \) ✓; at \( x = 4 \), \( 16 - 28 + 12 = 0 \) ✓.
2. Solve \( x^2 - 4 = 0 \).
Here \( b = 0 \), so this is a difference of squares: \( x^2 - 4 = (x - 2)(x + 2) \).
Setting each factor to zero gives \( x - 2 = 0 \) or \( x + 2 = 0 \).
The roots are \( x = \mathbf{2} \) and \( x = \mathbf{-2} \), often written \( x = \pm 2 \). (You could also just add \( 4 \) to both sides and take the square root: \( x^2 = 4 \Rightarrow x = \pm 2 \).)
3. Solve \( x^2 + 2x + 5 = 0 \).
Check the discriminant before reaching for factors: \( a = 1 \), \( b = 2 \), \( c = 5 \), so
\( D = b^2 - 4ac = 2^2 - 4(1)(5) = 4 - 20 = -16 \).
Because \( D < 0 \), there is no real root — the parabola \( y = x^2 + 2x + 5 \) sits entirely above the \( x \)-axis and never crosses it. (Its lowest point is the vertex at \( x = -\tfrac{b}{2a} = -1 \), where \( y = 4 \), already above zero.)