Quadratic Equations

Solving by Factoring #

If a product is zero, a factor is zero.

If two things multiply to zero then one of them must be zero

Two brackets multiply to 0, so one of them must be 0.

Each bracket has its own zero: x = 2 and x = 3 — two answers rather than one.

The Quadratic Formula #

Solves the ones that will not factor.

The formula solves any quadratic including the ones that will not factor

Read a, b and c straight off the equation in this order.

Here it is — plug in a, b and c, and both answers come out at once.

For $x^2-2x-3$x² − 2x − 3: $b^2-4ac$b² − 4ac is 16, its root 4, so $x=\frac{\left(2\pm 4\right)}{2}$x = (2 ± 4)/2 — that is 3 or −1.

The two answers are where the curve crosses the axis.

The Sum and Product of the Roots #

Both read off a, b and c without solving.

The sum and the product of the roots are read off a quadratic without solving it

A quadratic with roots $\alpha$α and $\beta$β is its leading number a times $\left(x-\alpha \right)\left(x-\beta \right)$(x − α)(x − β).

Match the x terms: $b=-a\left(\alpha +\beta \right)$b = −a(α + β), so the sum of the roots is $-\frac{b}{a}$−b/a.

Match the constants: $c=a\alpha \beta$c = aαβ, so the product of the roots is $\frac{c}{a}$c/a.

For $2x^2-10x+3$2x² − 10x + 3 the roots add to 5 and multiply to $\frac{3}{2}$3/2. Neither was found.

Check it: $x^2-5x+6$x² − 5x + 6 has roots 2 and 3, which add to 5 and multiply to 6.

The Discriminant and the Nature of the Roots #

Its sign counts the crossings.

The sign of the discriminant says how many times the curve meets the axis

Under the root sits $b^2-4ac$b² − 4ac — the discriminant. Its sign runs the whole show.

For $x^2-4$x² − 4 it is 16, above zero: the curve crosses the axis in two places.

For $x^2-2x+1$x² − 2x + 1 it is 4 − 4 = 0: one repeated root, and the curve just touches.

For $x^2+2$x² + 2 it is −8, below zero: no real root, and the curve never comes down.

Below zero is not the end of the story — a later stage opens it with new numbers.

Always Positive or Always Negative Quadratics #

No crossing, then the sign of the leading term.

A quadratic keeps one sign when its curve misses the axis, and a decides which sign

$x^2+x+3$x² + x + 3 never reaches the axis, so its value is positive whatever x you pick.

$b^2-4ac$b² − 4ac is 1 − 12 = −11. Below zero means no real root and no crossing.

Turn a to −1 and the curve opens downward, still missing the axis: always negative.

Two facts settle it: the discriminant below zero, then the sign of a.

Let the discriminant climb above zero and the curve crosses — the sign changes.

Completing the Square #

Rewrites it so the turning point shows.

Completing the square rewrites a quadratic so its turning point is visible

$x^2+4x$x² + 4x is a square with its corner missing. How big does that corner have to be?

Half of 4 is 2, so ${\left(x+2\right)}^2$(x + 2)² fits — but it brings a spare 4, so take that back off.

A square is never below 0, so this bottoms out at −4, where x = −2 empties it.

Same reading of ${\left(x-2\right)}^2-3$(x − 2)² − 3: the bracket empties at 2, so the low point is (2, −3).

Sketching from Vertex and Factored Form #

Each form hands you its own landmarks.

Each written form hands you its own landmarks: the vertex, or the crossings

$y={\left(x-2\right)}^2+1$y = (x − 2)² + 1 wears its vertex: the square is zero at x = 2 — lowest point (2, 1).

A minus in front flips it: $y=-{\left(x-2\right)}^2+4$y = −(x − 2)² + 4 opens downward, highest point (2, 4).

y = (x − 1)(x − 5) shows its roots: crossings at 1 and 5, and the turn midway, at x = 3.

y = −(x − 1)(x − 5) keeps the same crossings and turns over the top instead.

Deriving the Quadratic Formula #

Complete the square on the general case.

Completing the square on the general equation produces the formula itself

Start with any quadratic and divide through by a, so the $x^2$x² stands alone.

Move the constant across. The left side is now ready to become a square.

Half of $\frac{b}{a}$b/a is $\frac{b}{2a}$b/2a. The bracket smuggles in $\frac{b^2}{4a^2}$b²/4a², so add it on the right too.

Put the right side over one bottom, $4a^2$4a². It tidies into $b^2-4ac$b² − 4ac over $4a^2$4a².

Root both sides. The root of $4a^2$4a² is 2a, and a root always brings its $\pm$±.

Move $\frac{b}{2a}$b/2a across, and there it is. Every quadratic ever, solved at once.

Forming Quadratic Equations #

Area and product problems become quadratics.

A problem about area or product often turns into a quadratic

A rectangle is x wide and two longer than that.

Its area is 24, and multiplying out gives a quadratic to solve.

Solving a Quadratic Inequality #

Critical values first, then pick the side.

A quadratic inequality is solved by factoring for the critical values then choosing a side

Factor first. Each bracket empties at its own value: the critical values 1 and 4.

Sketch it. Between 1 and 4 the curve dips under the axis, so the value is negative.

So the answer is the stretch between the critical values: 1 < x < 4.

Turn the sign around and you want the curve above the axis — left of 1 it is.

$x^2-5x+4>0$x² − 5x + 4 > 0 is therefore two rays: x < 1 or x > 4.

A Linear Equation Paired with a Quadratic #

Substitute the line and a quadratic is left.

Substituting the linear equation into the quadratic leaves one quadratic in one unknown

$y=x^2$y = x² and y = x + 2 cross twice, so this pair has two solutions, not one.

Put the line in for y: $x^2=x+2$x² = x + 2, then gather everything on one side.

Factor, and both x values fall out: x = 2 and x = −1.

Each x needs its own y, so send both back through the line to finish the pairs.

The two answers are the two crossings: (2, 4) and (−1, 1).

When a Line Meets, Touches or Misses a Curve #

The discriminant of the combined equation decides.

Substituting a line into a curve gives a quadratic whose discriminant counts the meetings

The line y = 2x cuts $y=x^2$y = x² twice. At a meeting point the two y values agree.

Set them equal and gather: every meeting point is a root of $x^2-2x-k=0$x² − 2x − k = 0.

Its discriminant is 4 + 4k, and the number of roots is the number of meetings.

At k = −1 the discriminant is 0: one repeated root, and the line just touches.

At k = −3 it is −8, below zero: no real root, and the line misses altogether.

Above zero cuts, exactly zero touches, below zero misses. One test, three answers.

Fractional Equations #

Clear the denominator and a quadratic appears.

Clearing the denominators can turn a fraction equation into a quadratic

The unknown sits underneath. Multiply both sides by x and a quadratic appears.

Factor it and both answers fall out: x = 3 or x = −2.

Solving Equations with a Square Root #

Squaring can invent an answer, so check each one.

Squaring both sides can create an answer the original equation does not have, so each one must be checked

Square both sides to clear the root, then expand the bracket on the right.

Gather and factor: the squared equation offers x = 1 and x = 6.

Check x = 6 in the original equation: $\sqrt{9}$√9 is 3, and 6 − 3 is 3 as well.

Check x = 1 and it breaks: $\sqrt{4}$√4 is 2, while 1 − 3 comes to −2.

Squaring accepts 2 = −2, so the extra answer appears there — only x = 6 holds.

Solving a Rational Inequality #

The sign of the bottom is unknown, so test it.

A rational inequality is solved by testing the sign of each factor, never by multiplying up

Multiplying up is barred: x − 4 may be negative, and that turns the sign round.

Test the sign of each factor instead. Two matching signs divide to a positive.

So the quotient is positive outside the critical values: x < 1 or x > 4.

Or multiply by ${\left(x-4\right)}^2$(x − 4)², which is safe, and a quadratic inequality is left.

x = 4 is barred whatever the sign asks: the denominator is zero.