Use the x-intercept method to find all real solutions of the equation. 14x^3- 53x^2+ 41x-4--4x^3+ x^2

+ 1x + 4

Skip to content# Use the x-intercept method to find all real solutions of the equation.14x^3- 53x^2+ 41x-4–4x^3+ x^2 + 1x + 4

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Comments (10) on “Use the x-intercept method to find all real solutions of the equation.14x^3- 53x^2+ 41x-4–4x^3+ x^2 + 1x + 4”

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+ 1x + 4

the answer on edgen is graph c

-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 ..

0=0 Commutative Law

Step-by-step explanation:

Commutative Law, a+b=b+a

The solutions of the given equation in increasing order are

[tex]x=-\dfrac{1}{2},~0,~\dfrac{3}{2}.[/tex]

Step-by-step explanation: The given equation is

[tex]4x^3-5x=|4x|~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]

We are to solve the above equation and to list the solutions in increasing order.

We know that

[tex]|x|=a~~~~~~\Rightarrow a=x~~\textup{or}~~a=-x.[/tex]

So, from equation (i), we get

[tex]4x^3-5x=4x~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(iii)\\\\\textup{or}\\\\4x^3-5x=-4x~~~~~~~~~~~~~~~~~~~~~~~~~~~~(iv)[/tex]

Solving equation (iii), we get

[tex]4x^3-5x=4x\\\\\Rightarrow 4x^3-9x=0\\\\\Rightarrow x(4x^2-9)=0\\\\\Rightarrow x=0,~~4x^2-9=0~~\Rightarrow x^2=\dfrac{9}{4}~~\Rightarrow x=\pm\dfrac{3}{2}.[/tex]

So, solutions of equation (i) are [tex]x=0,~\dfrac{3}{2},~-\dfrac{3}{2}.[/tex]

And, solving equation (iv), we get

[tex]4x^3-5x=-4x\\\\\Rightarrow 4x^3-x=0\\\\\Rightarrow x(4x^2-1)=0\\\\\Rightarrow x=0,~~~4x^2-1=0~~\Rightarrow x^2=\dfrac{1}{4}~~~\Rightarrow x=\pm\dfrac{1}{2}.[/tex]

So, solutions of equation (ii) are [tex]x=0,~\dfrac{1}{2},~-\dfrac{1}{2}.[/tex]

We can see that [tex]x=\dfrac{1}{2}~~\textup{and}~~x=-\dfrac{3}{2}[/tex] does not satisfy equation (i).

For [tex]x=\dfrac{1}{2},[/tex] we have

[tex]L.H.S.=4\times\dfrac{1}{8}-5\times\dfrac{1}{2}=-2,\\\\R.H.S.=|4\times\dfrac{1}{2}|=2\neq L.H.S.[/tex]

Similarly, for [tex]x=-\dfrac{3}{2},[/tex] we have

[tex]L.H.S.=4\times -\dfrac{27}{8}+5\times\dfrac{3}{2}=-6,\\\\R.H.S.=|4\times -\dfrac{3}{2}|=6\neq L.H.S.[/tex]

Thus, the solutions of the given equation in increasing order are

[tex]x=-\dfrac{1}{2},~0,~\dfrac{3}{2}.[/tex]

-1/2, 0, 3/2

this should be the answer, hopefully helped

True

Step-by-step explanation: You can make x equal to anything and both sides would be equal to each other

[tex]sin(4x)= \dfrac{3}{4}[/tex]

[tex]4x=sin^-^1( \dfrac{3}{4}), \pi -sin^-^1( \dfrac{3}{4})[/tex]

Since sin is a periodic function, add the periodicity.

[tex]4x=2\pi n+sin^-^1 ( \dfrac{3}{4} ) ,nEZ [/tex]

[tex]4x=2\pi n-sin^-^1 ( \dfrac{3}{4} ) ,nEZ[/tex]

[tex]x= \dfrac{2\pi n+sin^-^1 ( \dfrac{3}{4} ) }{4}, nEZ[/tex]

[tex]x= \dfrac{2\pi n+ \pi -sin^-^1 ( \dfrac{3}{4} ) }{4}, nEZ[/tex]

4x – 3 + 2x = 33

6x = 36

x = 6

D. x = 6

14x^3-4x^3-53x^2-x^2+41x-x-4-4=0 10x^3-54x^2+40x-8=0 2(5x-2)(x^2+5+2)=0 Divide both sides by 2 (5x-2)(x^2+5+2)=0 5x-2=0 5x=2 X=2/5 X^2+5+2=0 5+- sqrt (-5)^2-4×(1×2)/2×1 X= 5+- sqrt 17/2 So answer is X= 5+- sqrt 17/2, 2/5

1 solution

Step-by-step explanation:

It is an equation of the first degree (the greatest power value of x = 1) so it will have one solution

4x+2=4x+4-2x

2x=2

X=1