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

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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

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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 $x=\frac{2\pm 4}{2}$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

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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

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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

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$x^2+x+3$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

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$x^2+4x$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

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$y={\left(x-2\right)}^2+1$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) crosses at 1 and 5. Midway, x = 3 is its axis of symmetry, through the turn.

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

At x = 0, y = −(0 − 1)(0 − 5) = −5: the curve crosses the y-axis there.

Deriving the Quadratic Formula #

Complete the square on the general case.

Completing the square on the general equation produces the formula itself

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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, a product or a path often turns into a quadratic, and its vertex is the greatest value

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A rectangle is x wide and two longer than that.

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

$h=10t-5t^2$h = 10t − 5t² is the height in meters of a ball thrown upward, t seconds later. It is highest halfway to landing, at t = 1.

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

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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

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$y=x^2$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

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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

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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.

Check each root against the denominator you cleared: x = 2 makes x − 2 zero, so only x = −2 solves it.

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

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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

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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.