Finding the Equation Revision Notes for Grade 11 NSC Matric Mathematics

Finding the Equation

When you know the roots (solutions) of a quadratic equation, you can work backwards to find the original equation. This is a useful skill that often appears in NSC exams.

What does "finding the equation" mean?

Finding the equation means determining a quadratic equation when you are given its roots or solutions. This is the reverse process of solving a quadratic equation.

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Think of it as working in reverse: instead of starting with an equation and finding the roots, you start with the roots and construct the equation.

The method for finding equations from roots

When you have two roots, follow this systematic approach:

Step 1: Write the roots as separate equations

If the roots are $p$ and $q$ , write:

$x = p$ or $x = q$

Step 2: Use additive inverses to get zero

Rearrange each equation to equal zero:

$x - p = 0$ or $x - q = 0$

Step 3: Write as a product of factors

Combine the factors:

$(x - p)(x - q) = 0$

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Key point: Notice that the signs in the brackets are opposite to the given roots.

Step 4: Expand the brackets

Multiply out the factors to get the standard form:

$x^2 - (p + q)x + pq = 0$

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Worked Example 1: Integer Roots

Question: Find an equation with roots 13 and -5.

Solution:

Step 1: Assign a variable and write roots as equations

$x = 13$ or $x = -5$

Step 2: Use additive inverses to get zero

$x - 13 = 0$ or $x + 5 = 0$

Step 3: Write as product of factors

$(x - 13)(x + 5) = 0$

Notice the signs in brackets are opposite to the roots.

Step 4: Expand the brackets

$(x - 13)(x + 5) = x^2 + 5x - 13x - 65 = x^2 - 8x - 65 = 0$

Answer: $x^2 - 8x - 65 = 0$

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Worked Example 2: Fractional Roots

Question: Find an equation with roots $-\frac{3}{2}$ and 4.

Solution:

Step 1: Write roots as equations

$x = 4$ or $x = -\frac{3}{2}$

Step 2: Use additive inverses

$x - 4 = 0$ or $x + \frac{3}{2} = 0$

Step 3: Clear the fraction

Multiply the second equation by 2:

$x - 4 = 0$ or $2x + 3 = 0$

Step 4: Write as product of factors

$(x - 4)(2x + 3) = 0$

Step 5: Expand the brackets

$(x - 4)(2x + 3) = 2x^2 + 3x - 8x - 12 = 2x^2 - 5x - 12 = 0$

Answer: $2x^2 - 5x - 12 = 0$

Multiple possible equations

An interesting mathematical property is that multiple different equations can have the same roots. This happens when equations are scalar multiples of each other.

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Important insight: If you multiply every term in a quadratic equation by the same non-zero constant, you get a different equation with the same roots.

For example, if $x^2 - 8x - 65 = 0$ has roots 13 and -5, then:

$2x^2 - 16x - 130 = 0$ (multiply by 2)
$-3x^2 + 24x + 195 = 0$ (multiply by -3)

These all have the same roots but are different equations.

Exam tips

Understanding common pitfalls and good practices will help you succeed in examinations.

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Exam Tips

Check your signs: The most common mistake is getting the signs wrong in the factors
Verify your answer: Substitute your roots back into your final equation to check
Simplify fractions: When dealing with fractional roots, multiply through to eliminate fractions
Show all steps: Examiners want to see your working, especially the factored form

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Key Points to Remember:

The signs in brackets are opposite to the given roots
Always start by writing the roots as separate equations equal to zero
The factored form $(x - p)(x - q) = 0$ is the key step
Multiple equations can have the same roots if they differ by a constant factor
Check your final answer by substituting the original roots back into the equation

Finding the Equation (Grade 11 NSC Matric Mathematics): Revision Notes