Reducing an equation to quadratic form
Some equations contain the same expression in several places. Give that whole expression a temporary name, solve the resulting quadratic in the temporary variable and then reverse the substitution
\[u=\text{repeated expression}\]
- Name the repeated expression Replace the complete repeated base, including all its terms, with u \(\displaystyle u=2x-1\)
- Solve the outer quadratic Solve in u before returning to the original variable \(\displaystyle u^2-5u+6=0\Rightarrow(u-2)(u-3)=0\)
- Substitute back Use each temporary value in the substitution and solve for x \(\displaystyle 2x-1=2\Rightarrow x=\frac{3}{2}\)
- Check restrictions Reject temporary or original values that cannot satisfy the substitution or the original equation \(\displaystyle u=x^2\Rightarrow u\ge0\)
Worked example
Solve the equation below
\[(x+1)^2-7(x+1)+12=0\]
- \(\displaystyle u=x+1\)Name the repeated expression shown in this step
- \(\displaystyle u^2-7u+12=0\Rightarrow(u-3)(u-4)=0\)Solve the resulting quadratic by factorising
- \(\displaystyle u=3\text{ or }u=4\)Apply the null factor law to obtain the temporary values
- \(\displaystyle x+1=3\text{ or }x+1=4\Rightarrow x=2\text{ or }x=3\)Substitute each temporary value back and solve for x
Common mistakes
Not reversing the substitution
✕\(\displaystyle u=5\Rightarrow x=5\)✓\(\displaystyle u=5\Rightarrow x^2+4=5\Rightarrow x=\pm1\)
A root found for u is not an x-value until it is substituted back into the definition of uKeeping an invalid temporary value
✕\(\displaystyle u=x^2,\quad u=-2\Rightarrow x=\pm\sqrt{-2}\)✓\(\displaystyle u=x^2,\quad u=-2\text{ is rejected because }u\ge0\)
A square cannot be negative, so a negative value of u is outside the substitution's domainSkipping the original-equation check
✕\(\displaystyle \sqrt{x+1}=x-1\Rightarrow x=0\text{ or }x=3\)✓\(\displaystyle \sqrt{x+1}=x-1\Rightarrow x=3\)
Squaring can introduce an extraneous candidate, so each result must be tested in the original equationDifficulty levels
introName the square of the input as the temporary value, solve the resulting quadratic and reject its negative temporary root
standardName a repeated linear expression, solve the temporary quadratic and reverse the substitution for every x-value
challengeHandle a repeated nonlinear expression and reject values outside its domain
examReduce, solve, back-substitute and check every candidate in the original equation
Questions about this topic
What should be substituted for u?+
Use the complete expression that repeats in the equation, including its brackets and all terms
Why must the temporary values be substituted back?+
The temporary equation gives values of u, not directly values of x. The definition of u is needed to find the corresponding x-values
Why can a value be rejected after solving?+
The substitution may restrict the possible values, or algebra such as squaring may create a candidate that does not satisfy the original equation
Can I use a calculator?+
No. This topic is practised without a calculator
Which year level and chapter is this?+
Year 11, VCE Units 1 & 2, Mathematical Methods, chapter 3: Quadratic functions
In this chapter
Factorise, solve and reconstruct a rational quadraticCalculate and interpret a quadratic discriminantTransform a quadratic and find its key featuresComplete the square and link equivalent formsDetermine a quadratic rule from exact informationSolve a quadratic exactly with the quadratic formulaConstruct a rational quadratic from a surd rootSolve an equation that reduces to quadratic formAnalyse signs and solve a quadratic inequalityIntersections of a line and a quadratic graphDetermine parameter values from a discriminant conditionFormulate and solve a quadratic problemAnalyse and optimise a quadratic model