An Equation to satisfy

$\frac{x^3 - 2x^2}{x^2}=5$ , x is not equal to 0

$x^3 - 2x^2 = 5x^2$

$x^3 - 7x^2 = 0$

$x^2(x - 7)$

$x=0 , 7$

reject 0 and hence, $x=7$

$(x^3 - 2x^2)/x^2$ 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 $x = 7$ .

$\frac {x^3 - 2x^2}{x^2} = 5$

$\frac {x^2(x - 2)}{x^2} = 5$

divide the nominator's x^2 with the denominator's x^2, so it becomes

$x - 2 = 5$

$x = 5 + 2$

$x = 7$

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

( \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