aishumurali wrote:
I tried solving this sum .
And i ended up getting:
(x^2 – 25)^2 = x^2 – 10x + 25
(x^2 – 25)^2 = (x-5)^2
I took square root on both sides and then i got this:
(x-5)(x+5) = (x-5)
x + 5 = 1
x = -4
What is wrong in the method?
First, there's a problem when you took the square root of both sides by simply removing the power of 2 from both sides of the equation.
For example, notice that (-3)^2 = (3)^2
If we remove the power of 2 from both sides of the equation, we get -3 = 3, which is incorrect.
In general, if k^2 = j^2, then it could be the case that k = j, or k = -j
So, once we have (x^2 – 25)^2 = (x - 5)^2, we must consider two possibilities: (x^2 – 25) = (x - 5) and (x^2 – 25) = -(x - 5)
Another problem has to do with taking (x-5)(x+5) = (x-5) and dividing both sides of the equation by (x - 5) to get x + 5 = 1
By doing this you may have accidentally divided both sides of an equation by 0.
Notice that x = 5 is a solution to the equation (x-5)(x+5) = (x-5)
Also notice that, when x = 5, (x-5) = 0, so when you divided by (x-5) you are accidentally dividing by 0.
More importantly, this resulted in missing one of the solutions (x = 5)
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Brent Hanneson - founder of Greenlight Test Prep