Week 3 · The roots and the coefficients: unknown coefficients, forming a quadratic equation from its roots or from the roots of another equation · 3 lessons
1
Finding an unknown coefficient from a condition on the roots: a given sum or product, one root given, or a complex root with its conjugate
Mathematics Sec 1 Term 1, Unit 1, Lesson 3
GoalUse the sum and the product of the roots to find an unknown coefficient.
New words
condition on the roots · a fact about the roots, such as their sum, product or one root being knownmultiplicative inverse · 1/L is the multiplicative inverse of L, so L × (1/L) = 1additive inverse · −L is the additive inverse of L, so L + (−L) = 0
Explanation
The sum −b/a and the product c/a link the roots to the coefficients, so a fact about the roots gives an equation for the unknown coefficient. If you are told the sum, put −b/a equal to it; for the product use c/a. If one root is given, substitute it into the equation, or divide c by that root to get the other root. A complex root brings its conjugate: when the coefficients are real, p + qi and p − qi are roots together, so the sum is 2p and the product is p² + q². Other conditions: one root double (or three times) the other means the roots are r and 2r (or 3r); opposite roots have sum 0; reciprocal roots have product 1. After finding the unknown, check by writing the equation and its roots. Never solve the equation first if the question says without solving.
What you are told
What to use
sum or product of the roots
−b/a or c/a equals the given number
one root is given
put it into the equation, or use the product
a complex root p + qi
the other root is p − qi (real coefficients)
one root is k times the other
call the roots r and kr
roots opposite / reciprocal
sum 0 / product 1
Worked examples
The product of the roots of 4x² − 7x + c = 0 is 3/4. Find c. Product = c/4 = 3/4, so c = 3. Check: 4x² − 7x + 3 = (4x − 3)(x − 1) has roots 3/4 and 1, with sum 7/4 ✓.
One root of x² + mx + 10 = 0 is −2. Find m and the other root. The other root = 10/(−2) = −5. Then the sum is −2 + (−5) = −7 = −m, so m = 7. Check: x² + 7x + 10 = (x + 2)(x + 5) ✓.
3 + 2i is a root of x² − 6x + c = 0 with c real. The other root is the conjugate 3 − 2i (their sum is 6 ✓). So c = (3 + 2i)(3 − 2i) = 9 + 4 = 13.
One root of 2x² − 8x + c = 0 is three times the other. Find c. Roots r and 3r: sum 4r = 4, so r = 1 and the roots are 1 and 3. Product 3 = c/2, so c = 6.
Class activity
Test the rules. Pick roots 2 and 5 and write x² − 7x + 10 = 0. Change the 10 to 12 and find the new roots. Does the sum stay 7? Does the product follow c/a?
Practice
1
The sum of the roots of x² + bx + 12 = 0 is −7. Find b and the roots.
−b = −7, so b = 7. x² + 7x + 12 = (x + 3)(x + 4): roots −3 and −4.
2
2 is a root of x² − 5x + c = 0. Find c and the other root.
4 − 10 + c = 0, so c = 6. The other root is 5 − 2 = 3 (and 2 × 3 = 6 ✓).
3
1 − 2i is a root of x² + bx + c = 0 with b and c real. Find b and c.
The other root is 1 + 2i. Sum 2 = −b, so b = −2. Product 1 + 4 = 5 = c.
4
One root of 3x² − 10x + (2k − 1) = 0 is the reciprocal of the other. Find k.
Product 1: (2k − 1)/3 = 1, so k = 2. The equation 3x² − 10x + 3 = (3x − 1)(x − 3) has roots 1/3 and 3 ✓.
5
The roots of x² − (k − 4)x + 5 = 0 are opposites of each other. Find k.
Sum 0: k − 4 = 0, so k = 4. The equation is x² + 5 = 0 with roots ±√5 i.
