Question 214·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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The solutions to $x^2 - 10x + 1 = 0$ are $r$ and $s$ , where $r < s$ . The solutions to $x^2 - 2x - 3 = 0$ are $t$ and $u$ , where $t < u$ . The solutions to $x^2 - Bx + C = 0$ are $rt + su$ and $ru + st$ . What is the value of $C$ ?

Your answer

For a monic quadratic, the constant term equals the product of its roots. When new roots are built from roots of earlier quadratics, multiply the new expressions and rewrite the result using the earlier roots’ sums and products. Square a root sum to find a sum of squares, then use Desmos for the arithmetic. Don’t confuse the new roots’ sum with their product.

Hints

Hint 1
In a monic quadratic, the coefficient of $x^2$ is $1$ , so the constant is the product of its roots. What product do the two expressions for the final equation’s roots give you?
Hint 2
The first two equations give you sums and products without finding individual roots. For $x^2+bx+c=0$ , the roots add to $-b$ and multiply to $c$ .
Hint 3
After multiplying the expressions for the new roots, collect terms containing $rs$ and terms containing $tu$ . To find $r^2+s^2$ , square $r+s$ and remove the extra $2rs$ .

Step-by-step

Use root sums and products
Step 1
Write the product the question asks for
The final equation starts with $x^2$ , so its constant term is the product of its roots. That makes $C=(rt+su)(ru+st)$ .
Step 2
Get the first roots’ sum and product
For $x^2+bx+c=0$ , the roots add to $-b$ and multiply to $c$ . In the first equation, $b=-10$ and $c=1$ , so $r+s=10,\qquad rs=1$ . You don’t need either root by itself.
Step 3
Get the second roots’ sum and product
Apply the same root rule to $x^2-2x-3=0$ : $t+u=2,\qquad tu=-3$ . Keep the minus sign on the product.
Step 4
Multiply the new roots
Multiply every term in the first parentheses by every term in the second: $C=r^2tu+rst^2+rsu^2+s^2tu$ . Keeping only the first and last products would leave out two terms.
Step 5
Group the terms you can evaluate
Factor out $tu$ from the first and last terms, and $rs$ from the middle terms: $C=tu(r^2+s^2)+rs(t^2+u^2)$ . Now the known products $tu$ and $rs$ appear.
Step 6
Find a rule for each sum of squares
Expand the square of $r+s$ , which contains an extra $2rs$ : $(r+s)^2=r^2+2rs+s^2$ . Subtract $2rs$ to get the sum of squares: $r^2+s^2=(r+s)^2-2rs$ . Square the sum, then subtract twice the product to get the sum of squares. The same rule works for $t$ and $u$ .
Step 7
Calculate the first sum of squares
Substitute $r+s=10$ and $rs=1$ : $r^2+s^2=10^2-2(1)$ . Type $10^2-2(1)$ in Desmos; it shows $98$ .
Step 8
Calculate the second sum of squares
Use the same rule with $t+u=2$ and $tu=-3$ : $t^2+u^2=2^2-2(-3)$ . Add $2^2-2(-3)$ in Desmos; it shows $10$ .
Step 9
Calculate the constant term
Substitute both sums of squares and both root products: $C=(-3)(98)+(1)(10)$ . Add $-3(98)+1(10)$ in Desmos; it shows $-284$ . That is the product of the final equation’s roots, so $C=-284$ . Grid in -284.

Lessons that teach this

SAT Advanced AlgebraIntermediateCoreConnect quadratic roots and coefficients

Question 214·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

Strategy

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Step-by-step

Lessons that teach this

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Strategy

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