To recover a quadratic from its roots α and β, use x² − (α + β)x + αβ = 0. The sum of the roots goes in with a negative sign, and the product is the constant term.
This lesson belongs to quadratic structure and discriminants and prepares you for questions about roots that are related to another equation’s roots.
Why does the sum and product rule work?
If the roots are α and β, the quadratic is (x − α)(x − β). Expanding gives x² − (α + β)x + αβ. So the coefficient of x is the negative of the sum, and the constant is the product.
Reading it the other way, ax² + bx + c = 0 has sum of roots −b/a and product c/a. You can find these two numbers without solving anything.
How do I build the equation?
- Find the sum of the roots and the product of the roots.
- Substitute into x² − (sum)x + (product) = 0.
- Clear any fractions by multiplying through, if integer coefficients are requested.
- Check by substituting one root back in.
Worked example
Part (a). Find a quadratic equation with integer coefficients whose roots are 3 and −1/2.
Sum = 3 − 1/2 = 5/2. Product = 3 × (−1/2) = −3/2.
x² − (5/2)x − 3/2 = 0. Multiply by 2: 2x² − 5x − 3 = 0.
Check: the discriminant is 25 + 24 = 49, so x = (5 ± 7)/4, giving 3 and −1/2. Correct.
Part (b). The roots of x² − 5x + 2 = 0 are α and β. Find the equation whose roots are α + 1 and β + 1.
From the given equation, α + β = 5 and αβ = 2.
New sum = (α + 1) + (β + 1) = 5 + 2 = 7.
New product = (α + 1)(β + 1) = αβ + (α + β) + 1 = 2 + 5 + 1 = 8.
The equation is x² − 7x + 8 = 0.
Check: the original roots are (5 ± √17)/2, so the new roots are (7 ± √17)/2. Their sum is 7 and their product is (49 − 17)/4 = 8.
The mistake to watch for
The usual slip is the sign of the sum.
Mistaken answer: roots 3 and −1/2, so x² + (5/2)x − 3/2 = 0
The student wrote the sum without the minus sign that the formula requires.
Test it: at x = 3, this gives 9 + 7.5 − 1.5 = 15, not 0. The correction is to write the template x² − ( )x + ( ) = 0 first and only then fill in the sum and product.
Check yourself
1. Find a quadratic with roots 4 and −7.
Show answer
Sum = −3, product = −28. So x² − (−3)x + (−28) = 0, which is x² + 3x − 28 = 0. Check at x = 4: 16 + 12 − 28 = 0.
2. Find an equation with integer coefficients whose roots are 2/3 and 1/2.
Show answer
Sum = 7/6, product = 1/3. So x² − (7/6)x + 1/3 = 0. Multiply by 6: 6x² − 7x + 2 = 0. Check at x = 1/2: 6/4 − 7/2 + 2 = 1.5 − 3.5 + 2 = 0.
3. The roots of 2x² − 6x + 1 = 0 are α and β. Find the equation whose roots are 2α and 2β.
Show answer
α + β = 3 and αβ = 1/2. New sum = 6, new product = 4 × 1/2 = 2. The equation is x² − 6x + 2 = 0. Check: the original roots are (3 ± √7)/2, so the doubled roots are 3 ± √7, with sum 6 and product 9 − 7 = 2.
Where this leads next
The same idea supports the factor theorem in polynomial factors and remainders. Within this module, go on to connect a turning point with a minimum value. The quadratic structure explorer shows factored and standard forms side by side, and the non-calculator working trainer practises exact fractions.
If you can solve equations but freeze when asked to build them, that is a pattern a teacher in online one-to-one Additional Mathematics tuition can work on with you directly.