Mathematics RU

Practice · Chapter 10

Quadratic equations

Quadratic equations: integer roots, fractional ones, roots with radicals, incomplete equations and ones with no roots. The discriminant, Vieta's formulas and every step explained.

How to solve it

A quadratic equation $ax^2 + bx + c = 0$ is solved by one formula, and the discriminant tells in advance how many roots there are. Incomplete equations are quicker without it: take out $x$ or isolate $x^2$.

Step by step

  1. Move everything to one side and expand the brackets to get $ax^2 + bx + c = 0$. If $a < 0$, multiplying the equation by $-1$ helps.
  2. An incomplete equation: with $c = 0$ take out $x$, $x(ax + b) = 0$; with $b = 0$ isolate $x^2 = -\frac{c}{a}$.
  3. Otherwise compute the discriminant $D = b^2 - 4ac$: $D > 0$ gives two roots, $D = 0$ one, $D < 0$ none.
  4. Use the quadratic formula; simplify the root of $D$ if you can and reduce the fraction.
  5. Check with Vieta's formulas: the roots add up to $-\frac{b}{a}$ and multiply to $\frac{c}{a}$.
The leading coefficient, of $x^2$. The coefficient of $x$, with its sign. The constant term. The discriminant: its sign tells how many roots there are. Example: $x^2 + 6x - 4 = 0$; $D = 36 + 16 = 52$, $\sqrt{52} = 2\sqrt{13}$; $x = \frac{-6 \pm 2\sqrt{13}}{2} = -3 \pm \sqrt{13}$.

Common mistakes

  • Losing the sign of a coefficient: in $x^2 - 5x + 6 = 0$ the coefficient is $b = -5$, so $-b = 5$.
  • Dividing only the root by $2$ instead of the whole sum: in $\frac{-6 \pm 2\sqrt{13}}{2}$ both $-6$ and $2\sqrt{13}$ are divided.
  • Dividing $x^2 - 6x = 0$ by $x$ and losing the root $x = 0$.
  • Writing roots with $\sqrt{-\ldots}$ when $D < 0$: there are no real roots.

Example