Question
If the roots of the equation x2+bx+c=0 differ by 1, find the relation between b and c.
Solution — Step by Step
Let the two roots be α and β. By Vieta’s formulas for x2+bx+c=0:
- Sum of roots: α+β=−b
- Product of roots: αβ=c
We are given that the roots differ by 1, so: ∣α−β∣=1, i.e., (α−β)2=1.
We use the identity:
(α−β)2=(α+β)2−4αβ
Substituting the Vieta’s expressions:
(α−β)2=(−b)2−4c=b2−4c
We know (α−β)2=1, so:
b2−4c=1
b2−4c=1
This is the required relation between b and c.
Take α=3, β=2 (differ by 1). Then:
- b=−(α+β)=−5
- c=αβ=6
Check: b2−4c=25−24=1 ✓
The equation would be x2−5x+6=0, which factors as (x−2)(x−3)=0 with roots 2 and 3 — indeed differing by 1.
Why This Works
The identity (α−β)2=(α+β)2−4αβ is derived from expanding (α+β)2=α2+2αβ+β2 and (α−β)2=α2−2αβ+β2. Subtracting gives the 4αβ term.
The discriminant Δ=b2−4ac of a quadratic ax2+bx+c=0 satisfies (α−β)2=Δ/a2. For our equation (where a=1), (α−β)2=b2−4c. This is why the discriminant determines the nature of roots — it literally measures the squared difference between them.
Alternative Method
If α−β=1 (assuming α>β), we have two equations:
α+β=−bandα−β=1
Adding: 2α=1−b, so α=(1−b)/2.
Subtracting: 2β=−b−1, so β=(−b−1)/2.
Now αβ=c:
4(1−b)(−b−1)=c⟹4−(1−b)(1+b)=c⟹4−(1−b2)=c
b2−1=4c⟹b2−4c=1✓
Same result, just a longer route.
Common Mistake
Students sometimes write ∣α−β∣=1 as α−β=1 and forget the absolute value, then proceed correctly. This is fine for the algebra (squaring removes the sign ambiguity). But a more subtle error is confusing b2−4c with b2+4c — the product of roots contributes with a minus sign in (α−β)2=(α+β)2−4αβ. Write out the identity explicitly before substituting to avoid sign errors.
This type of condition problem — where you’re given a relationship between the roots and asked to find a relationship between coefficients — always follows the same approach: (1) write Vieta’s formulas, (2) express the given condition in terms of sum/product, (3) substitute. The only identity you need to memorise is (α−β)2=(α+β)2−4αβ.