AMC 10 · 2003 · #5
Grade 8 algebraLet d and e denote the solutions of 2x2+3x−5=0. What is the value of (d−1)(e−1)?
Pick an answer.
AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: The equation $2x^{2}+3x-5=0$ has two solutions, called $d$ and $e$. Using those two solutions, find the value of the product $(d-1)(e-1)$.
Givens: $d$ and $e$ are the two solutions (roots) of $2x^{2}+3x-5=0$; In $2x^{2}+3x-5$ the leading coefficient is $a=2$, the middle coefficient is $b=3$, and the constant is $c=-5$; Answer choices: (A) $-\frac{5}{2}$, (B) $0$, (C) $3$, (D) $5$, (E) $6$
Unknowns: The value of $(d-1)(e-1)$
Understand
Restated: The equation $2x^{2}+3x-5=0$ has two solutions, called $d$ and $e$. Using those two solutions, find the value of the product $(d-1)(e-1)$.
Givens: $d$ and $e$ are the two solutions (roots) of $2x^{2}+3x-5=0$; In $2x^{2}+3x-5$ the leading coefficient is $a=2$, the middle coefficient is $b=3$, and the constant is $c=-5$; Answer choices: (A) $-\frac{5}{2}$, (B) $0$, (C) $3$, (D) $5$, (E) $6$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #16 Change Focus / Count the Complement, #6 Guess and Check
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 $2x^{2}+3x-5$ 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.
Execute — Answer: B
6.EE.A.3 Step 1 Expand the target product
- Multiply out $(d-1)(e-1)$ using the distributive property: $(d-1)(e-1)=de-d-e+1$.
- Group the two single terms as one sum: $de-(d+e)+1$.
- So the answer depends only on the product $de$ and the sum $d+e$ of the two roots — the individual values of $d$ and $e$ are not needed.
💡 Expanding first shows the answer only needs the roots' sum and product, so there is no reason to hunt down each root.
8.EE.C.7 Step 2 Read sum and product off the coefficients
- A quadratic with roots $d$ and $e$ and leading coefficient $2$ can be written as $2(x-d)(x-e)$.
- Expanding gives $2x^{2}-2(d+e)x+2de$.
- Matching this to $2x^{2}+3x-5$, the $x$-terms give $-2(d+e)=3$, so $d+e=-\frac{3}{2}$, and the constants give $2de=-5$, so $de=-\frac{5}{2}$.
💡 Writing the quadratic as its factored form makes the sum and product of the roots appear directly as the middle and constant coefficients.
7.NS.A.1 Step 3 Substitute and simplify
- Put $de=-\frac{5}{2}$ and $d+e=-\frac{3}{2}$ into $de-(d+e)+1$: $-\frac{5}{2}-\left(-\frac{3}{2}\right)+1=-\frac{5}{2}+\frac{3}{2}+1$.
- The two halves combine to $-\frac{2}{2}=-1$, and $-1+1=0$.
- So $(d-1)(e-1)=0$, which is choice (B).
💡 Once the sum and product are known, the answer is just a short piece of signed-fraction arithmetic.
6.EE.A.3 Multiply out $(d-1)(e-1)$ using the distributive property: $(d-1)(e-1)=de-d-e+1$ 8.EE.C.7 A quadratic with roots $d$ and $e$ and leading coefficient $2$ can be written as 7.NS.A.1 Put $de=-\frac{5}{2}$ and $d+e=-\frac{3}{2}$ into $de-(d+e)+1$: $-\frac{5}{2}-\l Review
Reasonableness: 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 $2x^{2}+3x-5$ 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).
Alternative: Factor the quadratic directly: $2x^{2}+3x-5=(2x+5)(x-1)$, whose roots are $x=-\frac{5}{2}$ and $x=1$. Then $(d-1)(e-1)=\left(-\frac{5}{2}-1\right)(1-1)=\left(-\frac{7}{2}\right)(0)=0$, the same answer (B) — and it shows that whichever root you call $d$ or $e$, the factor coming from the root $1$ zeroes out the product.
CCSS standards used (min grade 8)
6.EE.A.3Apply the properties of operations to generate equivalent expressions (Expanding $(d-1)(e-1)$ into $de-(d+e)+1$ so only the roots' sum and product are needed.)8.EE.C.7Solve linear equations in one variable (Matching $2(x-d)(x-e)$ to $2x^{2}+3x-5$ and solving $-2(d+e)=3$ and $2de=-5$ for the sum and product.)7.NS.A.1Apply and extend understanding of addition and subtraction to rational numbers (Combining $-\frac{5}{2}+\frac{3}{2}+1$ to reach $0$.)
⭐ 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.
⭐ 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.
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