How do you figure it out?
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Square both sides to solve quadratic:
x + 3 = (3 - x)²
--> multiply out
x + 3 = 9 - 6x + x²
--> set to zero (I'm going to switch sides)
x² - 6x - x + 9 - 3 = 0
--> collect like terms
x² - 7x + 6 = 0
We're in luck: 1*6 = 6 and 1 + 6 = 7...or actually: -1*-6 = +6 and -1 + -6 = -7
So that's what the factors look like:
(x - 1)(x - 6) = 0
--> solve now by setting each factor to zero
x - 1 = 0 --> x = 1
x - 6 = 0 --> x = 6
Now check to make sure that's fine to put into the square root:
x = 1: 1 + 3 = 4 <-- √4 = 2...we can check 3 - 1 = 2, CHECK
x = 6: 6 + 3 = 9 <-- √9 = 3...''__________" 3 - 6 = -3, WRONG
If we square both sides...then indeed -3² = 3² = +9, but NOT when taking the square root.
Therefore there is only one solution:
x = 1
x + 3 = (3 - x)²
--> multiply out
x + 3 = 9 - 6x + x²
--> set to zero (I'm going to switch sides)
x² - 6x - x + 9 - 3 = 0
--> collect like terms
x² - 7x + 6 = 0
We're in luck: 1*6 = 6 and 1 + 6 = 7...or actually: -1*-6 = +6 and -1 + -6 = -7
So that's what the factors look like:
(x - 1)(x - 6) = 0
--> solve now by setting each factor to zero
x - 1 = 0 --> x = 1
x - 6 = 0 --> x = 6
Now check to make sure that's fine to put into the square root:
x = 1: 1 + 3 = 4 <-- √4 = 2...we can check 3 - 1 = 2, CHECK
x = 6: 6 + 3 = 9 <-- √9 = 3...''__________" 3 - 6 = -3, WRONG
If we square both sides...then indeed -3² = 3² = +9, but NOT when taking the square root.
Therefore there is only one solution:
x = 1
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√(x+3) = 3-x (original equation)
[√(x+3)]² = (3-x)² (square both sides)
x+3 = (3-x)² (rewrite)
x+3 = x²-6x+9 (expand)
3 = x²-7x+9 (subtract x)
0 = x²-7x+6 (subtract 3)
0 = (x-6)(x+1) (factor)
x-6 = 0
x = 6
x+1 = 0
x = 1
Because we had to square a radical earlier, it is imperative that we check the solutions to see if one of them may be erroneous:
√[(1)+3] = 2
√(4) = 2
±2 = 2
2 = 2
True - the solution checks correctly
√[(6)+3] = 3-(6)
√(9) = -3
±3 = -3
3 = -3
3 ≠ -3
False - the solution does not check correctly
Solutions: x = 1
[√(x+3)]² = (3-x)² (square both sides)
x+3 = (3-x)² (rewrite)
x+3 = x²-6x+9 (expand)
3 = x²-7x+9 (subtract x)
0 = x²-7x+6 (subtract 3)
0 = (x-6)(x+1) (factor)
x-6 = 0
x = 6
x+1 = 0
x = 1
Because we had to square a radical earlier, it is imperative that we check the solutions to see if one of them may be erroneous:
√[(1)+3] = 2
√(4) = 2
±2 = 2
2 = 2
True - the solution checks correctly
√[(6)+3] = 3-(6)
√(9) = -3
±3 = -3
3 = -3
3 ≠ -3
False - the solution does not check correctly
Solutions: x = 1
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Start by squaring both sides, resulting in
x+3 = (3-x)^2
x+3 = 9 - 6x + x^2
then bring all of the terms to one side
x^2 - 7x + 6 = 0
then factor the equation to get
(x-6)(x-1) = 0
first let x-6 = 0
then, x = 6
BUT we have to take in account the square root
√(6 + 3) = 3 - 6
3 = -3
See? It doesn't work.
then let x-1 = 0
then, x = 1
If we but x = 1 into the equation,
√(1+3) = 3 - 1
2 = 2
therefore, x = 1
hope that helps!(:
x+3 = (3-x)^2
x+3 = 9 - 6x + x^2
then bring all of the terms to one side
x^2 - 7x + 6 = 0
then factor the equation to get
(x-6)(x-1) = 0
first let x-6 = 0
then, x = 6
BUT we have to take in account the square root
√(6 + 3) = 3 - 6
3 = -3
See? It doesn't work.
then let x-1 = 0
then, x = 1
If we but x = 1 into the equation,
√(1+3) = 3 - 1
2 = 2
therefore, x = 1
hope that helps!(:
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square both sides to get rid of the sqrt...
(x + 3) = (3 - x) * (3 - x)
(x + 3) = 9 - 3x - 3x + x^2
(x + 3) = 9 - 6x + x^2
now put everything to one part of equation
x^2 - 7x + 6
solve for x
(x - 1) (x - 6)
x = 1 and 6
(x + 3) = (3 - x) * (3 - x)
(x + 3) = 9 - 3x - 3x + x^2
(x + 3) = 9 - 6x + x^2
now put everything to one part of equation
x^2 - 7x + 6
solve for x
(x - 1) (x - 6)
x = 1 and 6
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√(x + 3) = 3 − x
[√(x + 3)]² = (3 − x)²
x+3=9-6x+x²
x²-7x+6=0
(x-1)(x-6)=0
x-1=0
x=1
x-6=0
x=6
[√(x + 3)]² = (3 − x)²
x+3=9-6x+x²
x²-7x+6=0
(x-1)(x-6)=0
x-1=0
x=1
x-6=0
x=6