Why proposed solution of an equation must be checked in the original equation?
Answers
We'll start with a couple simple exercises. As we go, remember that we must square the two sides of an equation, rather than the individual terms in those two sides. And when the instructions for an exercise say to check one's answers, this means that you need to show your check as part of your hand-in work.
If the instructions don't tell you that you must check your answers, check them anyway. At the very least, compare your solution with a graph on your graphing calculator.
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Solve the equation, and check.
\mathbf{\color{green}{\small{ \sqrt{\mathit{x} - 2\phantom{\big|}} = 5 }}}
x−2
∣
∣
∣
=5
On the left-hand side of this equation, I have a square root. By definition, this is positive. On the right-hand side, I've got a positive number. Since both sides are known positive, squaring won't introduce extraneous solutions. But I'll check my solution at the end, anyway, because the instructions require it.
First, I'll square both sides of the equation they've given me:
\small{ \left(\sqrt{x - 2\phantom{\big|}}\right)^2 = (5)^2 }(
x−2
∣
∣
∣
)
2
=(5)
2
x – 2 = 25
x = 27
Now, I need to show my check of this solution. When I plug 27 in for x in the left-hand side of the original equation, does the result simplify to the original right-hand side, "5"?
\small{ \sqrt{x - 2\phantom{\big|}} = \sqrt{(27) - 2\phantom{\big|}} }
x−2
∣
∣
∣
=
(27)−2
∣
∣
∣
\small{ = \sqrt{25\phantom{\big|}} = 5 }=
25
∣
∣
∣
=5
My solution checks, so my answer is:
x = 27
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algebra
Solve the equation, and check your answer.
\mathbf{\color{green}{\small{ \sqrt{\mathit{x} - 1\phantom{\big|}} = \mathit{x} - 7 }}}
x−1
∣
∣
∣
=x−7
The left-hand side of this equation is a square root. By definition, this will be positive. The right-hand side of the equation will sometimes be positive and sometimes be negative.
Because of this fact, my squaring of both sides of the equation will be an irreversible step. So I'd have been checking my solutions for this question, even if they hadn't told me to.
I'll treat the two sides of this equation as two functions, and graph them, so I have some idea what to expect. (I won't draw the graph or hand it is. This is for my own sense of confidence in my work.) I'll graph the two sides of the equation as:
\color{blue}{\small{ y_1 = \sqrt{x - 1\phantom{\big|}} }}y
1
=
x−1
∣
∣
∣
\color{green}{\small{ y_2 = x - 7 }}y
2
=x−7
y_1 = sqrt[x – 1] is blue arc, y_2 = x – 7 is green diagonal line; a red dot at (x, y) = (10, 3) marks the intersection point
The intersection point looks to be around x = 10, so I'll expect my solution to be at least close to that value.
To solve the equation properly (that is, algebraically), I'll start by squaring each side of the original equation:
\small{ \sqrt{x - 1\phantom{\big|}} = x - 7 }
x−1
∣
∣
∣
=x−7
\small{ \left(\sqrt{x - 1\phantom{\big|}}\right)^2 = (x - 7)^2 }(
x−1
∣
∣
∣
)
2
=(x−7)
2
x – 1 = (x – 7)(x – 7)
x – 1 = x2 – 14x + 49
x – 1 = x2 – 14x + 49
0 = x2 – 15x + 50
0 = (x – 5)(x – 10)
x = 5, 10
y1 = x – 1
y2 = x2 – 14x + 49
x = 5:
\small{ \mathrm{LHS: }\, \sqrt{(5) - 1\phantom{\big|}} = \sqrt{4\phantom{\big|}} = 2 }LHS:
(5)−1
∣
∣
∣
=
4
∣
∣
∣
=2
\small{ \mathrm{RHS: }\, (5) - 7 = -2 }RHS:(5)−7=−2
\small{ \mathrm{LHS} \neq \mathrm{RHS} }LHS
=RHS
So x = 5 is not a valid solution.
x = 10:
\small{ \mathrm{LHS: }\, \sqrt{(10) - 1\phantom{\big|}} = \sqrt{9\phantom{\big|}} = 3 }LHS:
(10)−1
∣
∣
∣
=
9
∣
∣
∣
=3
\small{ \mathrm{RHS: }\, (10) - 7 = 3 }RHS:(10)−7=3
\small{ \mathrm{LHS = RHS} }LHS=RHS
So x = 10 is a valid solution, and my answer is:
x = 10
Solve \mathbf{\color{green}{\small{ \sqrt{25 - \mathit{x}^2\phantom{\big|}} = 4 }}}
25−x
2
∣
∣
∣
=4
Square both sides:
\small{ \left(\sqrt{25 - x^2\phantom{\big|}}\right)^2 = (4)^2 }(
25−x
2
∣
∣
∣
)
2
=(4)
2
25 – x2 = 16
9 = x2
±3 = x
I have two possible solutions. The instructions didn't say to check them, but I'm going to, to be on the safe side:
x = –3:
\small{ \mathrm{LHS: }\, \sqrt{25 - (-3)^2 \phantom{\big|}} }LHS:
25−(−3)
2
∣
∣
∣
\small{ = \sqrt{25 - 9\phantom{\big|}} }=
25−9
∣
∣
∣
\small{ = \sqrt{16\phantom{\big|}} = 4 }=
16
∣
∣
∣
=4
RHS: 4
RHS = LHS
So this solution works. What about the other one?
x = 3:
\small{ \mathrm{LHS: }\, \sqrt{25 - (3)^2\phantom{\big|}} }LHS:
25−(3)
2
∣
∣
∣
\small{ = \sqrt{25 - 9\phantom{\big|}} }=
25−9
∣
∣
∣
\small{ = \sqrt{16\phantom{\big|}} = 4 }=
16
∣
∣
∣
=4
RHS: 4
RHS = LHS
So both solutions are valid, and my answer is:
x = ±3