What real value of x satisfies the equation
x 2 x 3 − 2 x 2 = 5 ?
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A mistake that is often made, is to forget that 0 cannot be a solution, because we end up dividing by 0.
This mistake was also made in a much harder question in this weeks set.
oh my god i can't understand everything.
0 it is a possible solution.
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no 0 is not a possible solution because 0/0 is not possible.
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For x^3-2x^2 = 5x^2, 0 it is a possible solution.
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@Manuel Ignacio López Quintero – Indeed, 0 is a possible solution for x^3-2x^2 = 5x^2, but not our question up there. If you substitute x=0 into the equation there, your answer is undefined, since no number can be divided by 0.
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@Chin Fong Wong – Ok, I understand it. Thanks.
( x 3 − 2 x 2 ) / x 2 simplifies to x − 2 , so we have the equation x − 2 = 5 . By adding 2 to both sides of the equation, we end up with x = 7 .
x 2 x 3 − 2 x 2 = 5
x 2 x 2 ( x − 2 ) = 5
divide the nominator's x^2 with the denominator's x^2, so it becomes
x − 2 = 5
x = 5 + 2
x = 7
As pointed out, you cannot simply "divide by x 2 ", unless you have verified that it is non-zero.
arigato gozamusi
it is not actually possible to divide the numerator with x^2, unless there was an assumption that x is not equal to zero.
x³-2x²/x² = 5 => x-2=5 => x=5+2 => x=7
Get the common of the expression on the left side then cancel out x2. Then transpose 2 to the right side,add up to 5, it will become 7..That's it!!!
First you should cross multiply to get x^3 - 2x^2 = 5x^2. Bring the 5x^2 to the other side through subtraction to get x^3 - 7x^2 = 0. Take x^2 out of both terms to get x^2(x - 7) = 0. x can equal either 0 or 7.
Cross multiplying, we get x^3-2x^2=5x^2. Therefore x^3=7x^2. Therefore x^1=7. Or x=7
Be careful with cross-multiplying. You might introduce an addition solution, like 0 in this case. This solution is incomplete.
( \frac{x^3- 2x^2}{x^2} ) = 5 x-2=5 x=5+2=7
make it poly nomial form and solve we get three solutio 0,0,7 so 7 is real solution
equation whichcan written as x^3/x^2 - 2x^2/x^2=5, ,x^1-2 =5, x=5+2, =7
Given eqn is x3−2x2x2=5 x3 = 7x2 so,x=7 satisfies the given equation
x^3-2x^2=5x^2; x^3-2x^2-5x^2=0 x^2(x-7)=0 either x=0 or x=7 but x can't be equal to 0 for real value of x hence x=7
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x 2 x 3 − 2 x 2 = 5 , x is not equal to 0
x 3 − 2 x 2 = 5 x 2
x 3 − 7 x 2 = 0
x 2 ( x − 7 )
x = 0 , 7
reject 0 and hence, x = 7