BTGModeratorVI wrote: ↑Mon Jun 22, 2020 6:21 am
The sum of all solutions for x in the equation x^2 – 8x + 21 = |x – 4| + 5 is equal to:
(A) –7
(B) 7
(C) 10
(D) 12
(E) 14
Answer:
D
Source: Manhattan prep
There are 3 steps to solving equations involving ABSOLUTE VALUE:
1. Apply the rule that says: If |x| = k, then x = k and/or x = -k
2. Solve the resulting equations
3. Plug solutions into original equation to check for extraneous roots
Given: x² – 8x + 21 = |x – 4|+ 5
Subtract 5 from both sides: x² – 8x + 16 = |x – 4|
Apply above
rule
We get two cases: x² – 8x + 16 = x – 4 and -(x² – 8x + 16) = x – 4
x² – 8x + 16 = x – 4
Add 4 to both sides: x² – 8x + 20 = x
Subtract x from both sides: x² – 9x + 20 = 0
Factor: (x - 5)(x - 4) = 0
So,
x = 5 or
x = 4
-(x² – 8x + 16) = x – 4
Simplify: -x² + 8x - 16 = x – 4
Add 4 to both sides: -x² + 8x - 12 = x
Subtract x from both sides: -x² + 7x - 12 = 0
Multiply both sides by -1 to get: x² - 7x + 12 = 0
Factor: (x - 3)(x - 4) = 0
So,
x = 3 or
x = 4
We have three potential solutions:
x = 5,
x = 4 and
x = 3
Now let's test for EXTRANEOUS ROOTS
x = 5
Plug into original equation to get:
5² – 8(
5) + 21 = |
5 – 4|+ 5
Evaluate: 6 = 6
WORKS! So,
x = 5 IS a valid solution
x = 4
Plug into original equation to get:
4² – 8(
4) + 21 = |
4 – 4|+ 5
Evaluate: 5 = 5
WORKS! So,
x = 4 is a valid solution
x = 3
Plug into original equation to get:
3² – 8(
3) + 21 = |
3 – 4|+ 5
Evaluate: 6 = 6
WORKS! So,
x = 3 is a valid solution
SUM of all solutions =
5 +
4 +
3 =
12
Answer: D
