Question 190·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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\frac{1}{s}+\frac{1}{s-p}=\frac{3}{p}

The equation above relates real numbers $p$ and $s$ , where $p\ne 0$ , $s\ne 0$ , and $s\ne p$ . What is the sum of all possible values of $s/p$ ?

A rational equation has variables in denominators, so keep track of values that make them zero. If the question asks for a ratio, replace that ratio with one variable and clear the fractions. For $ax^2+bx+c=0$ , the roots add to $-b/a$ . Use Desmos to check that both roots are real and allowed; the midpoint of the roots is only half their sum.

Hints

Hint 1
Because $p\ne0$ , you can let $x=s/p$ . Then $s=px$ and $s-p=p(x-1)$ . Substitute these into the equation so the ratio you need becomes the variable you solve for.
Hint 2
A restriction is a value that makes an original denominator zero. After you substitute $s=px$ , which values of $x$ make $px$ or $p(x-1)$ equal zero? Keep those values excluded when you clear the fractions.
Hint 3
The roots of a quadratic are the values that make it zero. If you need their sum, use $-b/a$ from $ax^2+bx+c=0$ rather than finding each root separately.

Step-by-step

Rewrite the ratio and use the root sum
Step 1
Make the requested ratio the variable
Since $p\ne0$ , let $x=\frac{s}{p}$ , so $s=px$ and $s-p=p(x-1)$ . Substitute both expressions into the given equation: $\frac{1}{px}+\frac{1}{p(x-1)}=\frac{3}{p}$ . Now the ratio being asked for is $x$ .
Step 2
Keep the original restrictions
The given $s\ne0$ means $px\ne0$ , so $x\ne0$ . The given $s\ne p$ means $p(x-1)\ne0$ , so $x\ne1$ . These are forbidden values, even after the denominators disappear.
Step 3
Remove the common factor of $p$
Because $p\ne0$ , multiply every term by $p$ : $\frac1x+\frac1{x-1}=3$ . The same nonzero factor cancels from each term.
Step 4
Clear the remaining denominators
Since $x\ne0,1$ , you can multiply every term by the common denominator $x(x-1)$ : $(x-1)+x=3x(x-1)$ . Multiply the $3$ on the right too; it is a term in the equation.
Step 5
Expand the right side
Expand the right side:
$(x-1)+x=3x^2-3x.$
Step 6
Combine the left side
Combine the left side:
$2x-1=3x^2-3x.$
Step 7
Write a quadratic in $x$
Subtract $2x-1$ from both sides: $0=3x^2-5x+1$ . The middle coefficient is $-5$ , including its minus sign.
Step 8
Check that there are two real roots
The discriminant $b^2-4ac$ tells you whether a quadratic has real roots. Here $a=3$ , $b=-5$ , and $c=1$ . Type $(-5)^2-4(3)(1)$ in Desmos; it shows $13$ . Because that is positive, the quadratic has two distinct real roots.
Step 9
Check that neither root is forbidden
Type $q(x)=3x^2-5x+1$ , then $q(0)$ and $q(1)$ . Desmos shows $1$ and $-1$ , respectively. A root must make $q(x)=0$ , so neither forbidden value is a root. Both real roots count.
Step 10
Find the sum, not the midpoint
For $ax^2+bx+c=0$ , the root-sum formula is $-b/a$ . Type $-(-5)/3$ on the next Desmos line; it shows $1.666\ldots$ , and its fraction button gives $\frac53$ . Both roots are allowed, so the sum of all possible values of $s/p$ is $\frac53$ . Choice D.

Question 190·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

Strategy

Hints

Step-by-step

Lessons that teach this

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Strategy

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Step-by-step

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