User talk:2603:7000:B500:5D4:6190:D593:84BB:ACC8

Subject of the quadratic equation

Steps

ax² + bx + c = 0

4a²x² + 4abx + 4ac = 0

4a²x² + 4abx = –4ac

4a²x² + 4abx + b² = b² – 4ac

(2ax + b)² = b² – 4ac

2ax + b = ±√b² – 4ac

2ax = –b ± √b² – 4ac

x = ⁠–b ± √b² – 4ac/2a⁠

How to make x the subject of the quadratic equation, i.e. derive the quadratic formula

• Multiply the left side by 4a, as the right side is zero.

• Add –4ac to both sides.

• Add b² to both sides to complete the square.

• Square-root both sides.

• Add –b to both sides.

• Divide both sides by 2a.

How to turn the quadratic equation, written as ax² + bx + c = 0, into (2ax + b)² = b² – 4ac

• Replace 4ac by b² on the left side.

• Keep (b² – 4ac) on the right side.

Theory

• By square completion, quadratic equations, written as ax² + bx + c = 0, become (2ax + b)² = b² – 4ac.

• This theory shows that (2ax + b)² is on the left side and that (b² – 4ac) is on the right side.

• With (2ax + b)² on the left side and (b² – 4ac) on the right side, many of those who solve quadratic equations, written as ax² + bx + c = 0, are experts, masters or professionals.

• x = ⁠–b ± √b² – 4ac/2a⁠ is the quadratic formula.

b² – 4ac

• (b² – 4ac) is the expression in the quadratic formula that determines how many solutions quadratic equations, written as ax² + bx + c = 0, have.

• When b² – 4ac > 0, there are two solutions.

• When b² – 4ac = 0, the only solution is x = –⁠b/2a⁠.

• When b² – 4ac < 0, there are no solutions.

• When (b² – 4ac) is a perfect square, there are two rational solutions.

• When (b² – 4ac) is not a perfect square, there are two irrational solutions.

Distances or differences between the solutions of quadratic equations

⁠√b² – 4ac/a⁠ is the distance or difference between the solutions of the quadratic equation, written as ax² + bx + c = 0. It determines how far apart the solutions of the quadratic equation are, even though it is positive due to b² > 4ac. When (b² – 4ac) is a perfect square, the distance or difference between the solutions of the quadratic equation is rational. When (b² – 4ac) is not a perfect square, the distance or difference between the solutions of the quadratic equation is irrational.

Quadratic expression forms

Vertex form conversion

ax² + bx + c = ⁠(2ax + b)² + 4ac – b²/4a⁠

Quadratic expression forms

• Quadratic expressions, written as (ax² + bx + c), become ⁠(2ax + b)² + 4ac – b²/4a⁠ when converted to their vertex forms.

• Quadratic expressions, written as (ax² + bx + c), can be factorized when (b² – 4ac) is a perfect square.

• When a > 0, quadratic expressions have minimum values, i.e. ax² + bx + c >= ⁠4ac – b²/4a⁠.

• When a < 0, quadratic expressions have maximum values, i.e. ax² + bx + c <= ⁠4ac – b²/4a⁠.

Parabolae

Regular form

y = ax² + bx + c

x-intercept count

• When b² > 4ac, there are two x-intercepts, i.e. parabolae cross or cut the x-axis twice.

• When b² = 4ac, the only x-intercept is x = –⁠b/2a⁠, i.e. parabolae touch or hit the x-axis only once.

• When b² < 4ac, there are no x-intercepts, i.e. parabolae never intersect the x-axis.

Distances or differences between the x-intercepts of parabolae

⁠√b² – 4ac/a⁠ is the distance or difference between the x-intercepts of the parabola, written as y = ax² + bx + c. It determines how far apart the x-intercepts of the parabola are, even though it is positive due to b² > 4ac. When (b² – 4ac) is a perfect square, the distance or difference between the x-intercepts of the parabola is rational. When (b² – 4ac) is not a perfect square, the distance or difference between the x-intercepts of the parabola is irrational.

y = ax² + bx + c where b² < 4ac

• When a > 0 but b² < 4ac, parabolae are entirely above or over the x-axis.

• When a < 0 and b² < 4ac, parabolae are entirely below or under the x-axis.

Vertex figures

• (–⁠b/2a⁠, ⁠4ac – b²/4a⁠) is the vertex figure lying on the parabola, written as y = ax² + bx + c.

• When a > 0, parabolae have minimum vertex figures.

• When a < 0, parabolae have maximum vertex figures.

• When b² = 4ac, the vertex figure is the only point of tangency at x = –⁠b/2a⁠.

Axes of symmetry

• x = –⁠b/2a⁠ is the axis of symmetry, which is in the middle and halfway between the x-intercepts of the parabola, written as y = ax² + bx + c where b² > 4ac.

• When b² = 4ac, the axis of symmetry is the only x-intercept of the parabola, though written as x = –⁠b/2a⁠.

Theories focusing on the coefficients of quadratic expressions

a > 0, b < 0, c > 0, b² > 4ac

a > 0, b < 0, 0 < c < ⁠b²/4a⁠

a > 0, b < –2√ac, c > 0

0 < a < ⁠b²/4c⁠, b < 0, c > 0

a > 0, b < –2√ac, 0 < c < ⁠b²/4a⁠

0 < a < ⁠b²/4c⁠, b < 0, 0 < c < ⁠b²/4a⁠

0 < a < ⁠b²/4c⁠, b < –2√ac, c > 0

0 < a < ⁠b²/4c⁠, b < –2√ac, 0 < c < ⁠b²/4a⁠

Quadratic function switch

Steps

ax² + bx + c = y

4a²x² + 4abx + 4ac = 4ay

4a²x² + 4abx = 4a(y – c)

4a²x² + 4abx + b² = b² + 4a(y – c)

(2ax + b)² = b² + 4a(y – c)

2ax + b = ±√b² + 4a(y – c)

2ax = –b ± √b² + 4a(y – c)

x = ⁠–b ± √b² + 4a(y – c)/2a⁠

Theory

• By square completion, quadratic functions, written as ax² + bx + c = y, become (2ax + b)² = b² + 4a(y – c).

• This theory shows that (2ax + b)² is on the left side and that {b² + 4a(y – c)} is on the right side.

• With (2ax + b)² on the left side and {b² + 4a(y – c)} on the right side, many of those who perform switches to quadratic functions, written as ax² + bx + c = y, are experts, masters or professionals.

• x = ⁠–b ± √b² + 4a(y – c)/2a⁠ is the switch to ax² + bx + c = y.

Tangents to the parabola at intersection points relative to the horizontal axis

Given y = ax² + bx + c where a > 0, b < 0, c > 0 and b² > 4ac

(2ax + b)√b² – 4ac + 2ay = 4ac – b² at (–⁠b + √b² – 4ac/2a⁠, 0)

2ay = (2ax + b)√b² – 4ac + 4ac – b² at (⁠–b + √b² – 4ac/2a⁠, 0)

Tangents to the parabola, written as y = ax² + bx + c, at intersection points relative to the x-axis where a > 0, b < 0, c > 0 and b² > 4ac

Theory

• (2ax + b)√b² – 4ac + 2ay = 4ac – b² is the tangent to the parabola, written as y = ax² + bx + c where a > 0, b < 0, c > 0 and b² > 4ac, at (–⁠b + √b² – 4ac/2a⁠, 0).

• 2ay = (2ax + b)√b² – 4ac + 4ac – b² is the tangent to the parabola, written as y = ax² + bx + c where a > 0, b < 0, c > 0 and b² > 4ac, at (⁠–b + √b² – 4ac/2a⁠, 0).

Ways to solve quadratic equations

• Factorization (for or used by experts, masters or professionals)

• Square completion (for or used by experts, masters or professionals)

• Quadratic formula (for or used by beginners)

• Graphing (for or used by beginners)

Whereas the quadratic formula is only for beginners, experts, masters or professionals solve quadratic equations by square completion. According to these, the solutions of the quadratic equation, written as ax² + bx + c = 0, are the x-intercepts of the parabola, written as y = ax² + bx + c.

Factorization

Factorization is a way to factorize the left side to solve quadratic equations. Experts, masters or professionals use the factorization method to solve quadratic equations. Some quadratic equations are unable to be solved by factorization.

Square completion

Square completion is a way to solve quadratic equations that are either hard to factorize to show integers or unable to be factorized. Experts, masters or professionals use the square completion method to solve quadratic equations. This method involves hard calculations, but it is best used when a = 1 and b is an even integer. It is the hardest quadratic-equation-solving method.

Quadratic formula

For quadratic equations that are either hard to factorize to show integers or unable to be factorized, the quadratic formula is required. Beginners use the quadratic formula to solve quadratic equations. This method consumes time when (b² – 4ac) is not written at first, but it is the easiest quadratic-equation-solving method.

Graphing

Parabola graphing is recommended when solving quadratic equations. Beginners use the graphing method to solve quadratic equations.

Ways to convert quadratic expressions

• Factorized form

• Vertex form

Factorized form

Quadratic expressions can be converted to their factorized forms by factorization. In other words, the form is provided by the factorization method.

Vertex form

Quadratic expressions can be converted to their vertex forms by square completion. In other words, the form is provided by the square completion method.

Vieta’s formula

x₁ + x₂ = –⁠b/a⁠

x₁x₂ = ⁠c/a⁠

(x₂ – x₁)² = (x₁ + x₂)² – 4x₁x₂

(x₂ – x₁)² = (–⁠b/a⁠)² – ⁠4c/a⁠

(x₂ – x₁)² = ⁠b² – 4ac/a²⁠

x₂ – x₁ = ±⁠√b² – 4ac/a⁠

⁠x₂/x₁⁠ = ⁠b² – 2ac ± b√b² – 4ac/2ac⁠

Value and change theories

First theory

When a > 0 and b² > 4ac, the quadratic expression:

• Decreases for x < –⁠b/2a⁠, i.e. before x = –⁠b/2a⁠.

• Has a minimum stationary value at x = –⁠b/2a⁠.

• Increases for x > –⁠b/2a⁠, i.e. after x = –⁠b/2a⁠.

• Is negative for –⁠b + √b² – 4ac/2a⁠ < x < ⁠–b + √b² – 4ac/2a⁠, i.e. after x = –⁠b + √b² – 4ac/2a⁠ but before x = ⁠–b + √b² – 4ac/2a⁠.

• Is zero at x = ⁠–b ± √b² – 4ac/2a⁠.

• Is positive for both x < –⁠b + √b² – 4ac/2a⁠, i.e. before x = –⁠b + √b² – 4ac/2a⁠, and x > ⁠–b + √b² – 4ac/2a⁠, i.e. after x = ⁠–b + √b² – 4ac/2a⁠.

Second theory

When a < 0 but b² > 4ac, the quadratic expression:

• Increases for x < –⁠b/2a⁠, i.e. before x = –⁠b/2a⁠.

• Has a maximum stationary value at x = –⁠b/2a⁠.

• Decreases for x > –⁠b/2a⁠, i.e. after x = –⁠b/2a⁠.

• Is positive for –⁠b + √b² – 4ac/2a⁠ < x < ⁠–b + √b² – 4ac/2a⁠, i.e. after x = –⁠b + √b² – 4ac/2a⁠ but before x = ⁠–b + √b² – 4ac/2a⁠.

• Is zero at x = ⁠–b ± √b² – 4ac/2a⁠.

• Is negative for both x < –⁠b + √b² – 4ac/2a⁠, i.e. before x = –⁠b + √b² – 4ac/2a⁠, and x > ⁠–b + √b² – 4ac/2a⁠, i.e. after x = ⁠–b + √b² – 4ac/2a⁠.

Third theory

When a > 0 and b² = 4ac, the quadratic expression:

• Decreases for x < –⁠b/2a⁠, i.e. before x = –⁠b/2a⁠.

• Has a minimum stationary value at x = –⁠b/2a⁠.

• Increases for x > –⁠b/2a⁠, i.e. after x = –⁠b/2a⁠.

• Is zero at x = –⁠b/2a⁠.

• Is positive for x ≠ –⁠b/2a⁠, i.e. every value of x except x = –⁠b/2a⁠.

Fourth theory

When a < 0 but b² = 4ac, the quadratic expression:

• Increases for x < –⁠b/2a⁠, i.e. before x = –⁠b/2a⁠.

• Has a maximum stationary value at x = –⁠b/2a⁠.

• Decreases for x > –⁠b/2a⁠, i.e. after x = –⁠b/2a⁠.

• Is zero at x = –⁠b/2a⁠.

• Is negative for x ≠ –⁠b/2a⁠, i.e. every value of x except x = –⁠b/2a⁠.

---