Question 162·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

Use $a=0$, $b=2$, and $c=1$. The first two are distinct and $c$ isn't zero, so a correct formula must work for this test. The original denominators also rule out $x=0$ and $x=2$.

Enter the chosen values and graph each side of the original equation as shown below. Click both intersections, where the two sides are equal. Desmos shows points near $(0.586,1)$ and $(3.414,1)$, so the allowed solutions for this test are about $0.586$ and $3.414$.

Type the plus-sign version of each of the three competing formulas. The new lines print $0$, $3$, and about $2.414$, respectively. None matches either intersection's $x$-value; $0$ is forbidden too. One failed proposed value rules out a formula that claims to give all solutions.

Type both signs of the remaining formula. Desmos prints about $3.414$ and $0.586$, matching both intersections. The other choices failed an allowed test, so the answer is $x=\dfrac{(a+b)+2c\pm\sqrt{(a-b)^2+4c^2}}{2}$. Choice A.

For the allowed values $x\ne a,b$, multiply both sides by $c(x-a)(x-b)$. Each fraction loses its denominator:

The two copies of $x$ add to $2x$. Combine like terms:

Distribute on the left and multiply the two factors on the right. Keep the minus signs attached to $a$ and $b$:

Subtract $2cx$ and add $c(a+b)$ on both sides so the equation equals zero:

Combine the two terms containing $x$:

This is a quadratic, an equation with an $x^2$ term. Its $x$ coefficient is $-(a+b+2c)$, not $-(a+b)$.

For $Ax^2+Bx+C=0$, the quadratic formula is $x=\frac{-B\pm\sqrt{B^2-4AC}}{2A}$. Here $A=1$, $B=-(a+b+2c)$, and $C=ab+c(a+b)$. Substitute each signed coefficient:

The discriminant is the expression under the square root. Expand it, then cancel the opposite $4c(a+b)$ terms:

The last line uses $(a+b)^2-4ab=(a-b)^2$. Don't drop $4c^2$ when the middle terms cancel.

Clearing fractions can sometimes create an extraneous solution: one that fits the new equation but makes an original denominator zero. Here, putting $x=a$ or $x=b$ into $c(2x-a-b)=(x-a)(x-b)$ makes the right side $0$, while the left side is $c(a-b)$ or $c(b-a)$. Both are nonzero because $c\ne0$ and $a\ne b$, so neither forbidden value is a root.

Put the simplified discriminant into the formula:

Since $c\ne0$, the expression under the square root is positive. The two signs give two distinct real values, and neither is forbidden. These are all possible values of $x$. Choice A.