The quadratic formula

One formula solves every quadratic equation: the quadratic formula. Factoring, or taking a square root (as in $\displaystyle {x^{2}}={9}$ or $\displaystyle {(x-1)^{2}}={4}$), is quicker when it works. But many trinomials cannot be factored with whole numbers: in $\displaystyle {x^{2}+2x-4}={0}$ no two whole numbers have the product $\displaystyle -4$ and the sum $\displaystyle 2$.

The solutions of the equation $\displaystyle {ax^{2}+bx+c}={0}$ (where $\displaystyle a\neq0$) are

The sign $\displaystyle \pm$ means two calculations: once with $\displaystyle +$ and once with $\displaystyle -$. Before substituting, write down $\displaystyle a$, $\displaystyle b$ and $\displaystyle c$ – each with its sign.

$\displaystyle {2x^{2}-5x-3}={0}$: $\displaystyle {a}={2}$, $\displaystyle {b}={-5}$, $\displaystyle {c}={-3}$.

$\displaystyle {x_{1}}={\frac{5+7}{4}}={3}$, $\displaystyle {x_{2}}={\frac{5-7}{4}}={-\frac{1}{2}}$. Check: $\displaystyle {2\cdot3^{2}-5\cdot3-3}={18-15-3}={0}$.

$\displaystyle {x^{2}+2x-4}={0}$: $\displaystyle {a}={1}$, $\displaystyle {b}={2}$, $\displaystyle {c}={-4}$.

The solutions are $\displaystyle {x_{1}}={-1+\sqrt{5}}$ and $\displaystyle {x_{2}}={-1-\sqrt{5}}$. We simplified $\displaystyle {\sqrt{20}}={\sqrt{4\cdot5}}={2\sqrt{5}}$, and divided both terms of the numerator by 2.

Solve with the quadratic formula.

The number of solutions depends on the expression under the root, $\displaystyle b^{2}-4ac$:

• positive – two solutions ($\displaystyle -b$ plus the root, and $\displaystyle -b$ minus it);
• zero – one solution, $\displaystyle {x}={\frac{-b}{2a}}$: $\displaystyle {x^{2}-6x+9}={0}$ gives $\displaystyle {36-36}={0}$, and $\displaystyle {x}={\frac{6}{2}}={3}$;
• negative – no solution: $\displaystyle {x^{2}+x+5}={0}$ gives $\displaystyle {1-20}={-19}$.

In the graph of $\displaystyle {y}={ax^{2}+bx+c}$ the solutions are where it meets the $\displaystyle x$-axis. Here $\displaystyle {y}={x^{2}-4x+3}$ (under the root $\displaystyle {16-12}={4}$) meets it twice, at 1 and at 3, and $\displaystyle {y}={x^{2}-4x+5}$ (dashed; under the root $\displaystyle {16-20}={-4}$) not at all. With zero under the root the vertex is on the $\displaystyle x$-axis.

Without solving: calculate $\displaystyle b^{2}-4ac$, and write how many solutions the equation has.

$\displaystyle {x^{2}+4x+5}={0}$

$\displaystyle {9x^{2}-12x+4}={0}$

$\displaystyle {2x^{2}-x-3}={0}$

The formula always works, but it is not always the shortest way. Look at the equation first:

• no $\displaystyle x$ term: $\displaystyle {x^{2}}={16}$ – take the root: $\displaystyle {x}={4}$ or $\displaystyle {x}={-4}$;
• no number term: $\displaystyle {x^{2}-6x}={0}$ – a common factor: $\displaystyle {x(x-6)}={0}$, so $\displaystyle {x}={0}$ or $\displaystyle {x}={6}$;
• a trinomial that factors easily: $\displaystyle {x^{2}-5x+6}={0}$ – $\displaystyle {(x-2)(x-3)}={0}$, so $\displaystyle {x}={2}$ or $\displaystyle {x}={3}$;
• otherwise – the formula: $\displaystyle {2x^{2}-3x-4}={0}$ gives $\displaystyle {x_{1,2}}={\frac{3\pm\sqrt{9+32}}{4}}={\frac{3\pm\sqrt{41}}{4}}$.

Solve each equation in the way that is shortest for it, and write which way you chose.

Solve the equation $\displaystyle {2x(x-1)}={x+5}$.
Bring everything to one side: simplify $\displaystyle 2x(x-1)-(x+5)$.
Solve, and check one solution by substitution.