AMC 10 · 2005 · #16
Grade 8 algebraThe quadratic equation x2+mx+n has roots twice those of x2+px+m, and none of m,n, and p is zero. What is the value of n/p?
Pick an answer.
AMC 10 2005 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 quadratic $x^2+mx+n$ has roots that are exactly twice the roots of $x^2+px+m$. None of $m$, $n$, $p$ is zero. Find the value of $n/p$.
Givens: First quadratic: $x^2+px+m$; Second quadratic: $x^2+mx+n$; The roots of $x^2+mx+n$ are twice the roots of $x^2+px+m$; None of $m$, $n$, $p$ equals $0$; Answer choices: (A) $1$, (B) $2$, (C) $4$, (D) $8$, (E) $16$
Unknowns: The value of the ratio $n/p$
Understand
Restated: The quadratic $x^2+mx+n$ has roots that are exactly twice the roots of $x^2+px+m$. None of $m$, $n$, $p$ is zero. Find the value of $n/p$.
Givens: First quadratic: $x^2+px+m$; Second quadratic: $x^2+mx+n$; The roots of $x^2+mx+n$ are twice the roots of $x^2+px+m$; None of $m$, $n$, $p$ equals $0$; Answer choices: (A) $1$, (B) $2$, (C) $4$, (D) $8$, (E) $16$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #13 Convert to Algebra, #7 Identify Subproblems
The roots are the hidden link between the two quadratics, so Tool #4 (Introduce a Variable) names them $r$ and $s$. A quadratic built from roots $r,s$ factors as $(x-r)(x-s)=x^2-(r+s)x+rs$, so each coefficient is just the sum or the product of the roots. Tool #13 (Convert to Algebra) turns 'twice the roots' into equations by matching coefficients. Tool #7 (Identify Subproblems) splits the work into two clean channels: what the sums of roots tell us, and separately what the products tell us. Each channel gives one link ($m=2p$ and $n=4m$), and chaining them answers $n/p$.
Execute — Answer: D
7.EE.A.1 Step 1 Name the roots of the first quadratic
- Let the roots of $x^2+px+m$ be $r$ and $s$.
- Any quadratic with roots $r$ and $s$ factors as $(x-r)(x-s)$, and expanding gives $x^2-(r+s)x+rs$.
- Matching this against $x^2+px+m$ term by term, the middle coefficient gives $r+s=-p$ and the constant gives $rs=m$.
💡 A quadratic is built from its roots, so its coefficients are the sum and product of those roots wearing a disguise.
7.EE.A.1 Step 2 Name the roots of the second quadratic
- The roots of $x^2+mx+n$ are twice those of the first, namely $2r$ and $2s$.
- The same factoring rule gives $(x-2r)(x-2s)=x^2-(2r+2s)x+(2r)(2s)$.
- Matching against $x^2+mx+n$: the middle coefficient gives $2r+2s=-m$, and the constant gives $(2r)(2s)=4rs=n$.
💡 Doubling each root doubles their sum and quadruples their product, and those two changes are exactly what the new coefficients record.
8.EE.C.8 Step 3 Bridge the two through the shared roots
- The roots $r,s$ are the bridge, so substitute the first quadratic's facts into the second.
- For the sums: $2r+2s=2(r+s)=2(-p)=-2p$, and this equals $-m$, so $-m=-2p$, giving $m=2p$.
- For the products: $n=4rs$, and since $rs=m$, this is $n=4m$.
💡 Handling the sum-relation and the product-relation separately keeps two moving parts from tangling into one messy equation.
8.EE.C.7 Step 4 Chain the links to get n/p
- Combine the two links.
- From $n=4m$ and $m=2p$, substitute to get $n=4(2p)=8p$.
- Since $p\neq 0$, divide both sides by $p$: $n/p=8$.
- That is choice (D).
💡 Once $m$ is tied to $p$ and $n$ is tied to $m$, one substitution ties $n$ straight to $p$.
7.EE.A.1 Let the roots of $x^2+px+m$ be $r$ and $s$. Any quadratic with roots $r$ and $s$ 7.EE.A.1 The roots of $x^2+mx+n$ are twice those of the first, namely $2r$ and $2s$. The 8.EE.C.8 The roots $r,s$ are the bridge, so substitute the first quadratic's facts into t 8.EE.C.7 Combine the two links. From $n=4m$ and $m=2p$, substitute to get $n=4(2p)=8p$. S Review
Reasonableness: The individual roots $r,s$ never appear in the final answer, so the ratio $n/p=8$ should hold for any valid pair of quadratics — a good sign the algebra is clean. Build a concrete example to confirm: take $p=8$. Then $m=2p=16$ and $n=8p=64$. The first quadratic is $x^2+8x+16=(x+4)^2$, with a double root at $-4$. Its roots doubled are $-8,-8$, and $(x+8)^2=x^2+16x+64$ does equal $x^2+mx+n$ with $m=16,\,n=64$. Indeed $n/p=64/8=8$, matching (D). The trap choice (E) $16$ comes from doubling twice (once for the sum, once again for the product) instead of using the product's factor of $4$; choice (C) $4$ comes from forgetting the extra factor of $2$ hidden in $m=2p$.
Alternative: Skip naming the roots and scale the whole polynomial. If $g(x)=x^2+px+m$ has roots $r,s$, then a polynomial whose roots are $2r,2s$ is $4\,g(x/2)=4\left(\tfrac{x^2}{4}+p\cdot\tfrac{x}{2}+m\right)=x^2+2px+4m$. This must be $x^2+mx+n$, so comparing coefficients gives $m=2p$ and $n=4m=8p$, hence $n/p=8$.
CCSS standards used (min grade 8)
7.EE.A.1Apply properties of operations to add, subtract, factor, and expand linear expressions with rational coefficients (Expanding $(x-r)(x-s)=x^2-(r+s)x+rs$ (and the doubled-root version) to read the sum and product of the roots directly off each quadratic's coefficients.)8.EE.C.8Analyze and solve pairs of simultaneous linear equations (Using the shared roots to link the two quadratics' coefficient equations, eliminating $r,s$ to obtain $m=2p$ and $n=4m$.)8.EE.C.7Solve linear equations in one variable (Chaining $n=4m$ with $m=2p$ and dividing by the nonzero $p$ to solve $n/p=8$.)
⭐ A quadratic's coefficients are just the sum and product of its roots in disguise, so naming the roots turns 'twice the roots' into simple equations you can chain together.
⭐ A quadratic's coefficients are just the sum and product of its roots in disguise, so naming the roots turns 'twice the roots' into simple equations you can chain together.
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