AMC 10 · 2003 · #5

Grade 8 algebra
vieta-formulasquadratic-equationspolynomial-roots convert-to-algebra ↑ Prerequisites: quadratic-equations
📏 Medium solution 💡 2 insights
Problem
The equation 2x²+3x-5=0 has two solutions, called d and e. Using those two solutions, find the value of the product (d-1)(e-1).

Pick an answer.

(A)
$-\frac{5}{2}$
(B)
0
(C)
3
(D)
5
(E)
6

AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

The target (d-1)(e-1) expands into de-(d+e)+1, so the only two facts it needs are the sum d+e and the product de of the roots — that is tool #16 (Change Focus): stop trying to find each root and focus instead on their sum and product. Tool #4 (Introduce a Variable) supplies those two facts cheaply: a quadratic 2x²+3x-5 with roots d,e must equal 2(x-d)(x-e), and expanding that and matching coefficients reads the sum and product straight off b and c. Tool #6 (Guess and Check) gives a fast sanity route: testing x=1 shows 2+3-5=0, so 1 is actually one of the roots, which forces a factor of (1-1)=0 into the product — a quick confirmation of the answer.

1STEP 1

Expand the target product

Expanding gives (d-1)(e-1)=de-d-e+1=de-(d+e)+1, so only the roots' product and sum matter.

(d-1)(e-1)=de-d-e+1=de-(d+e)+1
2STEP 2

Read sum and product off the coefficients

Matching 2(x-d)(x-e)=2x²-2(d+e)x+2de against 2x²+3x-5 gives d+e=-3/2 and de=-5/2.

2(x-d)(x-e)=2x²-2(d+e)x+2de=2x²+3x-5 → d+e=-3/2, de=-5/2
3STEP 3

Substitute and simplify

Putting de=-5/2 and d+e=-3/2 into de-(d+e)+1 gives -5/2+3/2+1, so -1+1=0 — choice (B).

-5/2-(-3/2)+1=-5/2+3/2+1=-1+1=0 → (B)
Answer
0
A product equals 0 exactly when one of its factors is 0, so a 0 answer is a strong hint that one root equals 1. Check it: plug x=1 into 2x²+3x-5 to get 2+3-5=0, so x=1 really is a root. Then that root contributes the factor (1-1)=0 to (d-1)(e-1), forcing the whole product to 0 — matching choice (B).
💡Key takeaway

To evaluate an expression in both roots of a quadratic, expand it into the roots' sum and product, which you can read straight off the coefficients — you rarely need to find the roots themselves.

  • Expand the target product
  • Read sum and product off the coefficients
  • Substitute and simplify