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Secondary 3 · Mathematics
Linear and quadratic equations · Equations

Quadratic Equations

A video game box has a square base with an area of 144 cm². How long is the side? You are looking for an $x$ such that $x^2 = 144$: this is a quadratic equation, because the unknown is squared. These equations can have two solutions, one, or none. There are two ways to solve them: factoring, when possible, and the quadratic formula, which always works.

General form

Graphically, $y = ax^2 + bx + c$ is a parabola, and the solutions are the points where it crosses the x-axis. Change $a$, $b$, and $c$ and see how many crossings there are:

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For each of the following quadratic equations, identify the coefficients $a$, $b$ and $c$ and state whether it is a complete or incomplete equation: a)…

For each of the following quadratic equations, identify the coefficients $a$, $b$ and $c$ and state whether it is a complete or incomplete equation:

  1. a) $x^2 - 5x + 6 = 0$
  2. b) $2x^2 - 32 = 0$
  3. c) $x^2 + 8x = 0$
  4. d) $3 - x^2 + 2x = 0$
Solve
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Check, without solving the equation, whether the proposed values of $x$ are solutions to the equation $x^2 - 4x + 3 = 0$: a) $x = 1$ b) $x = 0$ c) $x = 3$ d) $x = -1$

Check, without solving the equation, whether the proposed values of $x$ are solutions to the equation $x^2 - 4x + 3 = 0$:

  1. a) $x = 1$
  2. b) $x = 0$
  3. c) $x = 3$
  4. d) $x = -1$
Solve

Incomplete equations

They are solved without the formula, and faster:

x · (x + 3) = 0 x = 0 x + 3 = 0 → x = −3 If a · b = 0, then a = 0 or b = 0. (No other number times another gives 0.)
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Solve the following incomplete quadratic equations of the type $ax^2 + c = 0$. Remember that if the square root does not yield a real number, you must state so: a)…

Solve the following incomplete quadratic equations of the type $ax^2 + c = 0$. Remember that if the square root does not yield a real number, you must state so:

  1. a) $x^2 - 25 = 0$
  2. b) $3x^2 - 12 = 0$
  3. c) $x^2 + 16 = 0$
  4. d) $2x^2 = 50$
Solve
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Solve the following incomplete quadratic equations of the type $ax^2 + bx = 0$ by taking out a common factor: a) $x^2 - 7x = 0$ b) $2x^2 + 10x = 0$ c) $3x^2 = 6x$ d)…

Solve the following incomplete quadratic equations of the type $ax^2 + bx = 0$ by taking out a common factor:

  1. a) $x^2 - 7x = 0$
  2. b) $2x^2 + 10x = 0$
  3. c) $3x^2 = 6x$
  4. d) $x^2 + x = 0$
Solve
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The area of the square base of a video game storage box is $144\text{ cm}^2$. Calculate the length of the side of the base by setting up a quadratic equation.

The area of the square base of a video game storage box is $144\text{ cm}^2$. Calculate the length of the side of the base by setting up a quadratic equation.

x x Area 144 cm²
Solve

Solving by factoring

The idea behind incomplete equations works for many more equations: if you can write $ax^2 + bx + c$ as a product of two factors, the equation practically solves itself. It is the first method required by the Secondary Year 3 curriculum, and it is usually the fastest.

1. Common factor

When there is no constant term ($c = 0$), every term has an $x$: take it out. $\ 5x^2 - 20x = 0 \Rightarrow 5x\,(x - 4) = 0 \Rightarrow x = 0$ or $x = 4$.

2. Difference of squares

One square minus another square factors as the product of two conjugate binomials:

$$x^2 - a^2 = (x + a)(x - a)$$

So, $x^2 - 49 = 0 \Rightarrow (x + 7)(x - 7) = 0 \Rightarrow x = -7$ or $x = 7$. It also works with coefficients: $4x^2 - 9 = (2x + 3)(2x - 3) = 0 \Rightarrow x = \pm\frac{3}{2}$. Change $a$ and $b$ to see why $(a + b)(a - b) = a^2 - b^2$:

3. Trinomial $x^2 + bx + c$

It is the product of two binomials with a common term: $(x + p)(x + q) = x^2 + (p + q)\,x + p\,q$. Read backwards, to factor you look for two numbers $p$ and $q$:

The drawing shows $x^2 + 5x + 6 = (x + 2)(x + 3)$: the rectangle with sides $x + 2$ and $x + 3$ is split into a square $x^2$, two strips that add up to $5x$, and a rectangle of $2 \times 3 = 6$.

x²3x 2x6 x3 x2 x² + 5x + 6 = (x + 2)(x + 3) 2 + 3 = 5 → coefficient of x 2 · 3 = 6 → constant term

If you can't find the numbers (for example, in $x^2 - 5x - 1 = 0$ there are no integers that work), don't waste time: use the quadratic formula.

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Solve the following incomplete quadratic equations using the fastest method (taking out the common factor or rearranging directly for the square root): a)…

Solve the following incomplete quadratic equations using the fastest method (taking out the common factor or rearranging directly for the square root):

  1. a) $4x^2 - 100 = 0$
  2. b) $3x^2 + 15x = 0$
  3. c) $\frac{x^2}{2} - 8 = 0$
  4. d) $5x^2 = 2x$
Solve
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Given the quadratic equation $x^2 + kx + 12 = 0$, calculate the value of the coefficient $k$ knowing that one of the solutions to the equation is $x = 3$. Once $k$ is…

Given the quadratic equation $x^2 + kx + 12 = 0$, calculate the value of the coefficient $k$ knowing that one of the solutions to the equation is $x = 3$. Once $k$ is found, find the other solution of the equation.

Solve

The quadratic formula

For any complete equation, whether or not it can be factored:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

The $\pm$ gives two solutions: one by adding the root and another by subtracting it.

x = −b± b² − 4ac 2a change the sign of b discriminant Δ twice a ± → two solutions
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Apply the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to find the solutions to the following complete equations: a) $x^2 - 6x + 8 = 0$ b)…

Apply the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to find the solutions to the following complete equations:

  1. a) $x^2 - 6x + 8 = 0$
  2. b) $x^2 + x - 12 = 0$
Solve
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Solve the following quadratic equation by first simplifying the algebraic expressions through expanding the algebraic identities: $$(2x - 3)^2 - (x + 1)^2 = -5$$…

Solve the following quadratic equation by first simplifying the algebraic expressions through expanding the algebraic identities:

$$(2x - 3)^2 - (x + 1)^2 = -5$$

Remember that to apply the quadratic formula, you must first reduce the equation to the form $ax^2 + bx + c = 0$.

Solve

The discriminant: how many solutions?

What is inside the square root, $\Delta = b^2 - 4ac$, is called the discriminant and tells you how many solutions there are without solving:

Δ > 0two solutions Δ = 0one (double) Δ < 0none
$\Delta = b^2 - 4ac$ Solutions
$\Delta > 0$ Two distinct solutions
$\Delta = 0$ One solution (double): $x = -\frac{b}{2a}$
$\Delta < 0$ No real solution (the square root of a negative doesn't exist)
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Calculate the value of the discriminant $\Delta = b^2 - 4ac$ for each equation and state whether it has two real solutions, a repeated root, or no real solutions (you do…

Calculate the value of the discriminant $\Delta = b^2 - 4ac$ for each equation and state whether it has two real solutions, a repeated root, or no real solutions (you do not need to calculate the solutions):

  1. a) $x^2 - 4x + 4 = 0$
  2. b) $x^2 + 2x + 5 = 0$
  3. c) $x^2 - 5x - 1 = 0$
Solve
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Study the discriminant ($\Delta = b^2 - 4ac$) of the following equations and determine, without solving them, how many real solutions each one has. Justify your answer…

Study the discriminant ($\Delta = b^2 - 4ac$) of the following equations and determine, without solving them, how many real solutions each one has. Justify your answer in each case:

  1. a) $x^2 - 6x + 9 = 0$
  2. b) $2x^2 + 3x + 5 = 0$
  3. c) $3x^2 - 4x - 2 = 0$

What value should the constant term take in equation a) so that it suddenly has no solution?

Solve
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Given the quadratic equation $x^2 - (k-2)x + k = 0$, determine what values the parameter $k$ must take so that the equation has a single real solution (a repeated root).

Given the quadratic equation $x^2 - (k-2)x + k = 0$, determine what values the parameter $k$ must take so that the equation has a single real solution (a repeated root).

Solve

Extension: biquadratic equations

This type of equation is not part of the Secondary Year 3 curriculum, but it is useful for high school. In $x^4 - 13x^2 + 36 = 0$ only $x^4$ and $x^2$ appear. The trick is a change of variable: we call $t = x^2$ (so $x^4 = t^2$).

x⁴ − 13x² + 36 t = x² t² − 13t + 36 x = ±√t ±2, ±3
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Biquadratic equations can be solved by using a substitution $t = x^2$. Use this technique to find the four real solutions of the following equation: $$x^4 - 13x^2 + 36 =…

Biquadratic equations can be solved by using a substitution $t = x^2$. Use this technique to find the four real solutions of the following equation:

$$x^4 - 13x^2 + 36 = 0$$
Solve

Quadratic word problems

When solving a problem, discard the solutions that make no sense: a negative length, a number of people with decimals…

x(x + 1) = 156 x = 12 ✓ x = −13 ✗ not a natural number
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If we add three times an integer to its square, we get zero as a result. What numbers could these be? Set up the equation and find the possible solutions.

If we add three times an integer to its square, we get zero as a result. What numbers could these be? Set up the equation and find the possible solutions.

Solve
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The product of two consecutive natural numbers is equal to their sum increased by $19$ units. What are these numbers?

The product of two consecutive natural numbers is equal to their sum increased by $19$ units. What are these numbers?

Solve
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We want to put a wooden frame of constant width $x$ around a rectangular photograph measuring $10\text{ cm}$ by $15\text{ cm}$. If we know that the total area (photo…

We want to put a wooden frame of constant width $x$ around a rectangular photograph measuring $10\text{ cm}$ by $15\text{ cm}$. If we know that the total area (photo plus frame) must be $300\text{ cm}^2$, what should the width $x$ of the frame be?

x x 10 cm 15 cm Photo Frame A = 300 cm²
Solve

Summary

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