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algebra help

tlairmore

New member
Joined
Jan 22, 2010
Messages
2
Can someone help me? what is the solution to this problem, I am trying to help my daughter with her math. I can't make the square root sign, so I just indicated in parentheses.

(the square root of 6x)+9=2

doesn't x have to be a negative number, and even then, would the square root be negative? Is this an unsolveable problem?

tcl
 

mmm4444bot

Super Moderator
Staff member
Joined
Oct 6, 2005
Messages
10,153
The square-root sign always denotes the "principal" root; in other words, when a Real number has two square roots, the square-root symbol stands for the positive root.

The square root in this exercise equals negative seven. Since that is impossible, there is no solution for x in this equation.

I suspect that your daughter's teacher might have included this exercise to show what happens when we don't check our answers. Checking solutions is important with these types of equations.

Solving equations with radicals involves squaring both sides of the equation, and this process sometimes leads to false solutions.

Here's how the solution process goes.

\(\displaystyle \sqrt{6x} \;+\; 9 \;=\; 2\)

Isolate the radical.

\(\displaystyle \sqrt{6x} \;=\; -7\)

Square both sides.

\(\displaystyle 6x \;=\; 49\)

Solve for x.

\(\displaystyle x \;=\; \frac{49}{6}\)

Now, let's check this candidate by substituting it for x in the original equation.

\(\displaystyle \sqrt{6 \cdot \frac{49}{6}} \;+\; 9 \;=\; 2\)

\(\displaystyle \sqrt{49} \;+\; 9 \;=\; 2\)

\(\displaystyle 7 \;+\; 9 \;=\; 2\)

\(\displaystyle 16 \;=\; 2\)

Our candidate results in a false statement, so there is no solution.
 

tlairmore

New member
Joined
Jan 22, 2010
Messages
2
thank you very much
tcl
 
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