Question

$x^2(x-1)(x-4)-2x^3=-6x^2$
If x > 0, what is one possible solution to the equation above? (Hint: there are two real solutions)​

Answer 1

Answer :-

Simplifying the equation -

$\implies\sf x^2(x-1)(x-4)-2x^3= -6x^2$

$\implies\sf x^2 [ x( x - 4) - 1 ( x - 4 ) ] - 2x^3 = - 6x^2$

$\implies\sf x^2 [ x^2 - 4x - x + 4 ] - 2x^3 = -6x^2$

$\implies\sf x^2 [ x^2 - 5x + 4 ] - 2x^3 = -6x^2$

$\implies\sf x^4 - 5x^3 + 4x^2 - 2x^3 = -6x^2$

$\implies\sf x^4 - 5x^3 - 2x^3 + 4x^2 + 6x^2 = 0$

$\implies\sf x^4 - 7x^3 + 10x^2 = 0$

$\implies\sf \dfrac{x^4 - 7x^3 + 10x^2}{x^2} = \dfrac{0}{x^2}$

$\implies\sf \dfrac{x^4}{x^2} - \dfrac{7x^3}{x^2} + \dfrac{10x^2}{x^2} = 0$

$\implies\sf x^2 - 7x + 10 = 0$

Factorizing the equation -

$\implies\sf x^2 - 7x + 10 = 0$

By using middle term splitting method :-

$\implies\sf x^2 - 5x - 2x + 10 = 0$

$\implies\sf x( x - 5 ) - 2 ( x - 5 ) = 0$

$\implies\sf ( x - 2 )(x - 5 ) = 0$

Finding the roots -

$\implies\sf ( x - 2 )(x - 5 ) = 0$

$\sf x - 2 = 0$

$\implies\sf x = 2$

$\sf x - 5 = 0$

$\implies\sf x= 5$

Possible solutions of the given equation are 5 , 2.

Answer 2

${\large{\underbrace{\textsf{\textbf{\color{red}{➵}{\color{aqua}{\:\:SOLUTION\: :-}}}}}}}$

[Let's solve your equation step-by-step]

.x2(x−1)(x−4)−2x3=−6x2

x4−7x3+4x2=−6x2

Step 1: Subtract -6x^2 from both sides.

x4−7x3+4x2−−6x2=−6x2−−6x2

x4−7x3+10x2=0

Step 2: Factor left side of equation.

x2(x−2)(x−5)=0

Step 3: Set factors equal to 0.

x2=0 or x−2=0 or x−5=0

${\large{\underline{\textsf{\textbf{\color{red}{➵}{\color{blue}{\:\:ANSWER\: :-}}}}}}}$

x=0 or x=2 or x=5