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Construct a rational quadratic from a surd root

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Step 3 of 6

Conjugate roots and rational coefficients

For a quadratic with rational coefficients and nonzero leading coefficient, let an irrational root be p+q√r, where p and q are rational, q≠0, and r is a positive nonsquare integer. Its other root is p−q√r. Their sum and product are rational because the surd terms cancel

\[\alpha=3+2\sqrt{2},\quad\beta=3-2\sqrt{2}\]
  1. 1Keep the rational part The conjugate keeps the rational part and changes only the sign of the surd term \(\displaystyle 3+2\sqrt{2}\Rightarrow 3-2\sqrt{2}\)
  2. 2Use x minus each root Each root gives a factor formed by x minus that root \(\displaystyle x-\left(3+2\sqrt{2}\right)\text{ and }x-\left(3-2\sqrt{2}\right)\)
  3. 3Multiply as a difference of squares Multiplying the conjugate pair cancels the surd terms and leaves a rational expression \(\displaystyle \left(x-\left(3+2\sqrt{2}\right)\right)\left(x-\left(3-2\sqrt{2}\right)\right)=(x-3)^2-8\)
  4. 4Collect rational coefficients Expand the remaining rational expression and apply any leading scale \(\displaystyle (x-3)^2-8=x^2-6x+1\)
Year 11, VCE Units 1 & 2, Chapter 3, 4 difficulty levels, No calculatorBuild a worksheet

Conjugate roots and rational coefficients

For a quadratic with rational coefficients and nonzero leading coefficient, let an irrational root be p+q√r, where p and q are rational, q≠0, and r is a positive nonsquare integer. Its other root is p−q√r. Their sum and product are rational because the surd terms cancel

\[\alpha=p+q\sqrt{r},\quad\beta=p-q\sqrt{r}\]
  1. Keep the rational part The conjugate of p+q√r keeps p and changes only the sign of the surd term \(\displaystyle 2+\sqrt{3}\Rightarrow2-\sqrt{3}\)
  2. Use x minus each root A root r gives a factor x-r in the polynomial \(\displaystyle r=2+\sqrt{3}\Rightarrow x-r=x-\left(2+\sqrt{3}\right)\)
  3. Multiply as a difference of squares Conjugate factors cancel their surd terms when they are multiplied \(\displaystyle \left(x-\left(2+\sqrt{3}\right)\right)\left(x-\left(2-\sqrt{3}\right)\right)=(x-2)^2-3\)
  4. Collect rational coefficients After the surds cancel, expand the remaining rational expression and apply any leading scale \(\displaystyle 2\left[(x-2)^2-3\right]=2x^2-8x+2\)

Worked example

One root of a quadratic with rational coefficients is α = 1 + 2√3. The leading coefficient is 2. Construct the quadratic equation

\[\alpha=1+2\sqrt{3},\quad a=2\]
  1. \(\displaystyle \beta=1-2\sqrt{3}\)Keep the rational part and reverse only the sign of the surd term
  2. \(\displaystyle \alpha+\beta=(1+2\sqrt{3})+(1-2\sqrt{3})=2\)Add the conjugate roots so the radical terms cancel
  3. \(\displaystyle \alpha\beta=(1+2\sqrt{3})(1-2\sqrt{3})=1-(2\sqrt{3})^2=1-12=-11\)Multiply the conjugates as a difference of squares
  4. \(\displaystyle 2\left(x-\alpha\right)\left(x-\beta\right)=2\left[(x-1)^2-12\right]=2x^2-4x-22=0\)Form the two root factors, apply the leading coefficient and expand the rational result

Common mistakes

Using the same root twice
✕\(\displaystyle \left(x-\left(2+\sqrt{3}\right)\right)^2\)✓\(\displaystyle \left(x-\left(2+\sqrt{3}\right)\right)\left(x-\left(2-\sqrt{3}\right)\right)\)
A rational-coefficient quadratic requires the conjugate pair, not two copies of the stated root
Reversing the factor signs
✕\(\displaystyle \left(x+\left(2+\sqrt{3}\right)\right)\left(x+\left(2-\sqrt{3}\right)\right)\)✓\(\displaystyle \left(x-\left(2+\sqrt{3}\right)\right)\left(x-\left(2-\sqrt{3}\right)\right)\)
A root r gives the factor x-r, so using x+r changes the root's sign
Dropping the leading scale
✕\(\displaystyle \left(x-\left(1+\sqrt{2}\right)\right)\left(x-\left(1-\sqrt{2}\right)\right)\)✓\(\displaystyle 3\left(x-\left(1+\sqrt{2}\right)\right)\left(x-\left(1-\sqrt{2}\right)\right)\)
The leading scale changes the quadratic's coefficients and must remain in the factor product

Difficulty levels

introUse an integer-centre root with a unit surd coefficient to construct a rule
standardUse rational root components or a non-unit leading scale to recover the exact conjugate and coefficients
challengeUse a non-unit leading scale with exact rational root components to recover the conjugate and quadratic coefficients
examSelect the conjugate factor construction and fill the resulting exact coefficients

Questions about this topic

What is the conjugate of p + q√r?+

For rational p and q with q≠0 and a positive nonsquare integer r, p+q√r is irrational and its conjugate is p−q√r. Keep p and change only the surd sign. The rational-coefficient condition is needed to conclude that this is the other quadratic root

Why do conjugate roots give rational coefficients?+

Their sum and product are rational because the surd terms cancel. With leading coefficient a, expanding the factors gives ax² − a(root sum)x + a(root product), so the coefficients are rational

Why is the factor for a root r written as x - r?+

Substituting x = r into x - r gives zero, so x - r is the factor associated with the root r

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