Chapter 3 · Complex Numbers and Equations · Concept 2
3-11 Signs of Two Solutions
Without solving anything, you can decide whether both answers are positive, both negative, or one of each. It takes three small conditions — and the difficulty is never finding them, it is putting them together.
Guiding question
What has to be true about a, b and c for both roots to be positive?
Learning outcome. Use the relationship between roots and coefficients to study the signs of the solutions.
Three conditions — or one
When the signs of the two solutions are known, write down the conditions and find where they all hold at the same time.
(i) two different positive solutions
D > 0, α + β > 0, αβ > 0
(ii) two different negative solutions
D > 0, α + β < 0, αβ > 0
(iii) solutions with different signs
αβ < 0 — and that is all
No D condition is needed here. The book says so, and the reason is worth knowing: if the product of the roots is negative then they cannot be an imaginary pair, so D is bigger than zero automatically.
Notice what is the same in (i) and (ii). Both need D > 0 and αβ > 0 — the product being positive only says the two roots have the SAME sign. It is the sum that then says which sign. So the three conditions are really: two real roots, same sign, and then which sign.
Each condition allows a stretch of the m-line. The answer is where all three stretches overlap — not where any one of them holds. Drag m and watch the ticks and crosses change.
the book's Warm Up
x2 − 2mx + m + 6 = 0, wanted: two different negative solutions
D = 4m² − 4(m + 6) = 56.00
α + β = 2m = -8.00
αβ = m + 6 = 2.00
All three hold — so the roots really are two different negatives.
the roots are -0.258 and -7.742
The green strip is the answer: −6 < m < −2.
x² − 2mx + m + 6 = 0 has two different negative solutions. Find the range of m.
- 1
Write the three conditions down BEFORE any algebra.
D > 0, α + β < 0, αβ > 0Every mark in this question comes from having all three. Writing them first means none can be forgotten.
The mistake the book prints.
From α + β < 0 people write −2m < 0 and get m > 0. But α + β is −b/a, and b is already −2m, so the sum is +2m. The minus in the formula and the minus in b cancel each other.
Work with a partner. One of you draws the three ranges on a number line while the other works them out. Then pick a number inside your final answer and actually solve the equation with it — the two roots should have the signs the question asked for.
In a battery factory the efficiency factor x depends on a design setting m:
x2 + (m + 1)x + m + 4 = 0
The equation has two different real solutions. Find the range of m in each case.
You have now spent five lessons finding things out about an equation without solving it — how many roots, their sum, their product, their signs. Why is that worth doing?
Think about what a designer actually needs to know.
Losing the minus in the sum.
If b = −2m then α + β = +2m, not −2m. The book prints this as its Common Mistake, and it changes the whole answer.
Forgetting D > 0.
The sum and product say nothing unless the roots are real. Without D > 0, "two different negative solutions" can be satisfied by two imaginary ones on paper.
Taking a union instead of an overlap.
All three conditions must hold TOGETHER. Draw them on one number line and take only the part covered by all three.
Adding a D condition to the "different signs" case.
It is not wrong, but it is wasted work — αβ < 0 already forces D > 0. The book points this out in the margin.
Not checking the answer.
Pick one number inside your final range and solve the equation with it. If the two roots do not have the signs you were asked for, something went wrong.
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