Lesson 6: Solving Quadratic Equations. Factoring and QF
Lesson 6: Solving Quadratic Equations. Factoring and QF
Solve quadratic equations by factoring and with the quadratic formula.
Solving quadratic equations requires a good understanding of and proficiency with a number of concepts. A quadratic equation is an equation of the form ax2 + bx + c = 0. This is a standard form equation. A quadratic equation can also be recorded in the factored form a(x – r)(x – s) = 0, where r and s are the roots of the equation.
In order to determine the solutions (roots or x-intercepts of the parabola) of the quadratic equation and when the equation is given in a standard form, we can either try to factor the equation or use the Quadratic Formula.
The method of solving quadratic equations by factoring depends whether the standard form equation is given as a simple or as a complex trinomial and whether the factors could be easily identified.
If the a value of the equation is a = 1 and the equation is x2 + bx + c = 0, then x1 + x2 = -b and (x1)(x2) = c
However, if a is not 0 or 1, then the equation could be solved by factoring using the decomposition method (watch the video lesson to see how).
Sometimes factors are not very “nice” numbers. In such cases it is difficult to factor and the Quadratic Formula should be used instead.
Before setting up the Quadratic Formula it is always a good idea to check whether the equation is at all solvable. To check how many possible solutions the quadratic equation has we use the Discriminant.
The Discriminant is an expression that is part of the Quadratic Formula. The note above contains all possible scenarios for when the Discriminant is positive, negative or 0.
Once the quadratic equation has been solved, the solutions can be used to graph the quadratic function or a statement based on the real life problem could be made. For example, if we are talking about a kicked ball, then the x-intercepts represent the time when the ball touches the ground as it was kicked and when it landed.
Key ideas: Solving Quadratic Equations. Factoring and the Quadratic Formula
To solve ax2 + bx + c = 0, first get 0 on one side. Then:
- Factor if you can: common factor, simple trinomial, complex trinomial, difference of squares (a2 − b2 = (a − b)(a + b)) or perfect square. Then set each factor equal to 0.
- Otherwise use the quadratic formula: x = [−b ± √(b2 − 4ac)] / (2a).
- The discriminant D = b2 − 4ac tells you how many real roots there are before you solve: D > 0 two roots, D = 0 one root, D < 0 no real roots.
The roots of the equation are the zeros (x-intercepts) of the parabola y = ax2 + bx + c.
Worked examples
Example 1
Solve x2 − 5x − 14 = 0.
- Find two numbers that multiply to −14 and add to −5: −7 and 2.
- (x − 7)(x + 2) = 0.
- x = 7 or x = −2
Example 2
Solve 6x2 + x − 2 = 0.
- Complex trinomial: (2x − 1)(3x + 2) = 0. Check: 6x2 + 4x − 3x − 2 ✓
- x = ½ or x = −⅔
Example 3
Solve 2x2 − 4x − 3 = 0.
- It does not factor nicely: a = 2, b = −4, c = −3.
- x = [4 ± √(16 + 24)] / 4 = (4 ± √40) / 4 = (2 ± √10) / 2.
- x ≈ 2.58 or x ≈ −0.58
Example 4
How many real roots does 3x2 + 2x + 5 = 0 have?
- D = 22 − 4(3)(5) = 4 − 60 = −56.
- D < 0, so there are no real roots: the parabola never crosses the x-axis.
Common mistakes
- Factoring before moving every term to one side (x2 = 5x needs to become x2 − 5x = 0).
- Dividing both sides by x and losing the root x = 0.
- Sign errors with −b in the formula when b is negative.
- Writing −b ± √(…) over 2a but only dividing part of the numerator by 2a.
Questions students ask
Should I factor or use the formula?
Try factoring first; it is faster. If you cannot find the factors within a minute, use the formula: it always works.
What does “no real roots” look like on a graph?
The parabola is completely above or below the x-axis, so it has no x-intercepts.
Check your understanding
Practise with the Grade 10 quizzes. You’ll see right away if each answer is correct.