Answer:
-1/2 , 0 , 3/2
Step-by-step explanation:
Given equation is:
[tex]4x^3-5x = |4x|[/tex]
We know that [tex]|x|=a\\The\ solution\ will\ be:\\x=a\ and\ x=-a\\[/tex]
So, from given equation,we will get two solutions:
[tex]4x^3-5x = 4x\\4x^3-5x-4x=0\\4x^3-9x=0\\x(4x^2-9) = 0\\x = 0\\and\\4x^2-9 = 0\\4x^2=9\\x^2 = \frac{9}{4} \\\sqrt{x^2}=\sqrt{\frac{9}{4} }\\[/tex]
x= ±√3/2 , 0
and
[tex]4x^3-5x = -4x\\4x^3-5x+4x=0\\4x^3-x=0\\x(4x^2-1) = 0\\x = 0\\and\\4x^2-1 = 0\\4x^2=1\\x^2 = \frac{1}{4} \\\sqrt{x^2}=\sqrt{\frac{1}{4} }[/tex]
x= ±1/2 , 0
We can check that 1/2 and -3/2 do not satisfy the given equation.
[tex]4x^3-5x = |4x|\\Put\ x=1/2\\4(\frac{1}{2})^3 - 5(\frac{1}{2}) = |4 * \frac{1}{2}|\\ 4 * (\frac{1}{8)} - \frac{5}{2} = |2|\\ -2 = 2\\Put\ x=-\frac{3}{2} \\4(\frac{-3}{2})^3 - 5(\frac{-3}{2}) = |4 * \frac{-3}{2}|\\-6 = 6\\[/tex]
So, 1/2 and -3/2 will not be the part of the solution ..
So, the solutions in increasing order are:
-1/2 , 0 , 3/2 ..
Answer:
[tex]-\frac{1}{2},0,\frac{3}{2}[/tex]
Step-by-step explanation:
We are given that an equation
[tex]4x^3-5x=\mid x\mid[/tex]
We have to find the solution of given equation and arrange the solution in increasing order.
[tex]4x^3-5x=4x[/tex] when x >0
and [tex]4x^3-5x=-4x[/tex] when x < 0
because [tex]\mid x\mid =x when x > 0 [/tex]
=-x when x < 0
[tex]4x^3-5x-4x=0[/tex]
[tex]4x^3-9x=0[/tex]
[tex]x(4x^2-9)=0[/tex]
[tex]x(2x+3)(2x-3)=0[/tex]
Using identity [tex]a^2-b^2=(a+b)(a-b)[/tex]
[tex]x=0,2x+3=0,2x-3=0[/tex]
[tex]2x=3\implies x=\frac{3}{2}=1.5[/tex]
[tex]2x=-3 \implies x=-\frac{3}{2}=-1.5[/tex]
[tex]4x^3-5x=-4x=0[/tex]
[tex]4x^3-5x+4x=0[/tex]
[tex]4x^3-x=0[/tex]
[tex]x(4x^2-1)=0[/tex]
[tex]x(2x+1)(2x-1)=0[/tex]
[tex]x=0,2x+1=0[/tex]
[tex]2x-1=0[/tex]
[tex]2x-1=0[/tex]
[tex]2x=1 \ilmplies x=\frac{1}{2}=0.5[/tex]
[tex]2x+1=0[/tex]
[tex]2x=-1 \implies x=-\frac{1}{2}=-0.5[/tex]
When we substitute x=[tex]\frac{1}{2}[/tex]
[tex]4(\frac{1}{2})^3-\frac{5}{2}=\frac{1}{2}-\frac{5}{2}=\frac{1-5}{2}=-2[/tex]
[tex]\mid 4(\frac{1}{2})\mid=2[/tex]
[tex]-2\neq 2[/tex]
Hence, [tex]\frac{1}{2}[/tex] is a not solution of given equation.
When substitute [tex]x=\frac{-3}{2}[/tex]
[tex]4(\frac{-3}{2})^3+\frac{15}{2}=\frac{-27}{2}+\frac{15}{2}=\frac{-27+15}{2}=-6[/tex]
[tex]\mid 4(-\frac{3}{2}\mid=6[/tex]
[tex]-6\neq 6[/tex]
Hence, [tex]\frac{-3}{2}[/tex] is not a solution of given equation.
Substitute x=[tex]-\frac{1}{2}[/tex] in the given equation
[tex]4(-\frac{1}{2})^3+\frac{5}{2}=-\frac{1}{2}+\frac{5}{2}=2[/tex]
[tex]\mid 4(-\frac{1}{2})\mid=2[/tex]
[tex]2=2[/tex]
Hence, [tex]-\frac{1}{2}[/tex] is a solution of given equation.
Substitute [tex]x=\frac{3}{2}[/tex] in the given equation
[tex]4(\frac{3}{2})^3-\frac{15}{2}=\frac{27-15}{2}=6[/tex]
[tex]\mid 4(\frac{3}{2})\mid =6[/tex]
[tex]6=6[/tex]
Hence, [tex]\frac{3}{2}[/tex] is a solution of given equation.
Answer:[tex]-\frac{1}{2},0,\frac{3}{2}[/tex]
