Question 1027·Hard·Math

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$\lvert x-4\rvert = \sqrt{9-x}$

What is the product of all real solutions to the equation above?

An equation with both an absolute value and a square root can hide more than one real solution. Graph the original sides in Desmos and click every intersection to see how many there are. Then square both sides to get a quadratic and use its coefficients for the exact product. Multiplying rounded graph coordinates can give the wrong choice.

Hints

Hint 1
An intersection is a point where two graphs have the same height. Graph each side of the given equation separately. How many places do they meet, and which coordinate of each point gives a solution?
Hint 2
A square root and an absolute value are both nonnegative. That makes squaring both sides safe wherever the square root is defined. What equation remains after you square?
Hint 3
The roots of a quadratic are its solutions. For $ax^2+bx+c=0$ , their product is $c/a$ . Put your squared equation in that form before using the rule.

Step-by-step

Graph the solutions, then find their exact product
Step 1
Find every solution on the original graphs
Type $y=|x-4|$ and $y=\sqrt{9-x}$ . An intersection is where the two sides are equal. Click both intersections: Desmos shows about $(1.209,2.791)$ and $(5.791,1.791)$ . Their $x$ -coordinates are the two solutions, but these decimals are rounded.
Step 2
Square to get an exact equation
The square root is defined when $x\le9$ . There, it and the absolute value are both nonnegative, so squaring cannot hide a sign mismatch. Square the whole equation:
$(x-4)^2=9-x.$
Step 3
Expand the squared expression
Multiply $(x-4)$ by itself, keeping both middle terms:
$x^2-4x-4x+16=9-x.$
Combine the two $x$ terms:
$x^2-8x+16=9-x.$
Step 4
Put the quadratic equal to zero
To use the root-product rule, get zero on one side. Add $x$ to both sides:
$x^2-7x+16=9.$
Subtract $9$ from both sides:
$x^2-7x+7=0.$
Step 5
Read the exact product from the coefficients
For $ax^2+bx+c=0$ , the product of the roots is $c/a$ . Here $a=1$ and $c=7$ , so the product of all the real solutions is $7/1=7$ . Choice C.

Question 1027·Hard·Math

Strategy

Hints

Step-by-step

Lessons that teach this

Study Hints

Strategy

Hints

Step-by-step

Lessons that teach this