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A1.8 · Solve quadratic equations and compare strategies

Learn to solve quadratic equations and compare strategies through clear examples and targeted practice.

Ontario Grade 11 Mathematics

Quadratic Functions

Choosing and comparing strategies for Grade 11

A quadratic equation contains a squared variable, and solving it means finding the variable values that make the two sides equal. There is more than one useful strategy. Some equations factor quickly. Others are easier to solve by completing the square or by using the quadratic formula. A graph can show where the solutions lie and can help check an answer. In this lesson, you will practise these approaches and consider when each one is useful.

What you will learn

1. Prerequisite bridge: what makes an equation quadratic?

An equation says that two expressions have the same value. A solution is a value of the variable that makes the equation true. For example, substituting a possible value into both sides lets you check whether it works.
A quadratic equation can be written in the form ax2+bx+c=0ax^2+bx+c=0, where aa, bb, and cc are numbers and a≠0a\ne 0. The highest power of the variable is 2. The expression x2x^2 is the squared term. The equation is often arranged with zero on one side before you solve it.
You may already know how to factor an expression such as x2+5x+6x^2+5x+6 into two brackets. Factoring reverses expansion. You may also know the zero-product property: if two factors multiply to zero, at least one factor must be zero. These ideas help solve some quadratic equations.
The solutions of a quadratic equation are also called its roots. On a graph of y=ax2+bx+cy=ax^2+bx+c, a real solution is an xx-value where the graph crosses or touches the horizontal axis. The horizontal axis is the line y=0y=0.
ax2+bx+c=0,a≠0ax^2+bx+c=0,\quad a\ne 0

2. Three algebraic strategies

Factoring is often the quickest strategy when the quadratic expression breaks into simple factors. After writing the equation as a product equal to zero, set each factor equal to zero. This works because a product is zero only when at least one factor is zero. Factoring may take more thought when the coefficients are not simple.
Completing the square rewrites a quadratic so that part of it is a perfect square. A perfect square is an expression such as (x+3)2(x+3)^2. To keep an equation balanced, whatever you add to one side must also be added to the other. This method can make the solutions visible even when factoring is difficult, but it requires careful handling of the terms.
The quadratic formula gives solutions for any quadratic equation already written in standard form. The values of aa, bb, and cc are read from the equation, including their signs. This method is dependable, though substituting and simplifying can take longer than easy factoring.
These strategies solve the same equation, so they should give the same solutions. Choose by looking at the equation. Try factoring when a product is easy to find. Consider completing the square when you want a squared expression. Use the formula when factoring is not clear or when you want a consistent algebraic method.
A graph offers another view. The graph of a quadratic is a parabola, a U-shaped or upside-down U-shaped curve. Its horizontal-axis intersections represent real solutions. A graph or graphing technology can show approximate solutions, but a rounded estimate may not be exact. Use algebra or substitution to confirm exact answers when possible.
x=−b±b2−4ac2ax=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

3. Guided example: solve and compare

Suppose a rectangular garden has a length that is 5 metres more than its width. Its area is 36 square metres. If the width is xx metres, the area relationship becomes x(x+5)=36x(x+5)=36. This is a quadratic equation. First, arrange it with zero on one side.
This equation is a good candidate for factoring because the constant term and middle term can be matched by a pair of integers. The formula also works. Comparing both approaches shows why it is useful to inspect an equation before choosing a method.
x(x+5)=36x(x+5)=36

4. Apply, check, and choose

In an application, the equation may have more than one mathematical solution. Return to the situation and decide which values make sense. A length cannot be negative, for example. State units when the context uses measurements.
A careful comparison includes accuracy as well as speed. Factoring can be short, but guessing factors carelessly can lead to errors. Completing the square has clear steps, but a missed balance adjustment changes the equation. The formula has a reliable structure, but sign errors when identifying aa, bb, or cc are common. A graph is helpful for seeing the number and approximate location of real solutions, but it may not show exact values.
After solving, substitute each answer into the original equation. If both sides have the same value, the answer passes the check. For a contextual problem, also check whether the answer fits the described quantities. A solution that fails the context is not an acceptable answer to that application.

Worked example

Garden dimensions: two algebraic methods

A rectangular garden has length 5 metres greater than its width and area 36 square metres. Find its dimensions by solving x(x+5)=36x(x+5)=36, then compare factoring with the quadratic formula.
  1. Write standard form
    Expand the left side and subtract 36 from both sides. Keeping zero on one side makes the equation ready for either strategy.
    x2+5x−36=0x^2+5x-36=0
  2. Factor
    Find two numbers with product −36-36 and sum 55. The numbers 99 and −4-4 work, so write the quadratic as a product. By the zero-product property, set each factor equal to zero.
    (x+9)(x−4)=0(x+9)(x-4)=0
  3. Find the roots
    Solving each simple equation gives the two mathematical roots. The negative root cannot be a width, so keep the positive value for the garden. x=-9 or x=4
  4. Compare with the formula
    For the standard-form equation, a=1a=1, b=5b=5, and c=−36c=-36. Substituting these values gives the same two roots. Here, factoring is shorter. The formula is still useful when a convenient factor pair is not obvious.
    x=−5±52−4(1)(−36)2(1)=−5±132x=\frac{-5\pm\sqrt{5^2-4(1)(-36)}}{2(1)}=\frac{-5\pm13}{2}
  5. Interpret and check
    The width is 4 metres, so the length is 9 metres. Their product is 36 square metres, as required. This check also confirms that the positive root fits the situation.
    4(4+5)=364(4+5)=36
Answer: The garden is 4 metres wide and 9 metres long.
Check: The dimensions are positive, differ by 5 metres, and have area 36 square metres.

Common mistakes and how to avoid them

Forgetting to move every term to one side before factoring.
Correction: Arrange the equation so one side is zero, then check that the other side is equivalent to the original expression.
Setting a sum of factors equal to zero instead of setting each factor equal to zero.
Correction: Use the zero-product property only after the equation is written as factors multiplied together and equal to zero.
Using the wrong sign for a coefficient in the quadratic formula.
Correction: Compare the equation term by term with ax2+bx+c=0ax^2+bx+c=0. Include any negative signs when identifying the coefficients.
Keeping a negative solution in a measurement context without checking it.
Correction: Check each root against the original situation. A negative length or width does not describe the garden.
Treating a graph's rounded intersection as an exact solution.
Correction: Use the graph for a visual estimate, then verify with substitution or an algebraic method.

Lesson summary

Check your understanding

Question 1

Which strategy is usually most efficient for x2−7x+12=0x^2-7x+12=0?
  1. Factoring, because the expression has simple integer factors.
  2. The quadratic formula, because factoring is impossible.
  3. Graphing only, because an exact solution cannot be found algebraically.
  4. correctIndex
Show answer and explanation
Factoring, because the expression has simple integer factors.
The expression factors as (x−3)(x−4)(x-3)(x-4), so factoring gives the exact roots efficiently. The formula could also work, but it is not necessary here.

Question 2

A graph of a quadratic meets the horizontal axis at x=−2x=-2 and x=5x=5. What do these values represent?
  1. The yy-intercepts of the graph.
  2. The real solutions of the related quadratic equation.
  3. The maximum and minimum values of the graph.
  4. correctIndex
Show answer and explanation
The real solutions of the related quadratic equation.
Horizontal-axis intersections have y=0y=0. Their xx-coordinates are the real solutions of the related equation.

Question 3

When should you reject a mathematical root in a contextual problem?
  1. When it does not make sense for the quantity described.
  2. Whenever the root is negative, in every possible problem.
  3. Whenever it came from the quadratic formula.
  4. correctIndex
Show answer and explanation
When it does not make sense for the quantity described.
A root is rejected only when the context rules it out. A negative number may be valid in some contexts, but it cannot represent a garden width.

Key terms

Quadratic equation
An equation whose highest power of the variable is 2.
Solution or root
A value of the variable that makes an equation true.
Factoring
Rewriting an expression as a product of simpler expressions.
Zero-product property
If a product equals zero, at least one of its factors must equal zero.
Completing the square
Rewriting a quadratic equation to make a perfect-square expression.
Parabola
The U-shaped or upside-down U-shaped graph of a quadratic function.

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About this lesson

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Mathematics (MCF3M), expectation A1.8. It is a study resource, not an official curriculum publication.

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