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Welcome to A-Maths!

Today in this blog, I will solve an SAT math problem. This problem belongs to a Big exponent equation. Most students think that a big exponent problem is a difficult task. But you will see here inside this lesson, how we can solve this Question easily by using basic Algebraic rules.

Let's start the solution

So, The question is

x³(x²-5)= -4x

In order to solve this equation first multiply x³ with (x²-5), we will get x⁵-5x³, and -4x will remain the same.

x⁵ -5x³ = -4x

Now in the next step x⁵-5x³ will remain the same but this -4x will go on the other side of the Equation and -4x will change to plus 4x.

x⁵ - 5x³ + 4x =0

We can see that x is common in all values.

We will get common x from all values. So we will get 

x(x⁴-5x²+4)=0.

We can see clearly that this equation is a quadratic equation and factorizable equation.

Now we will factorize this equation. We can see that "4x" is the "c" of this quadratic equation.  It has a plus sign. It means that "-5x²" which is b of this quadratic equation will break in the same sign. So, it will be broken into -x²-4x².

x(x⁴-x²-4x²+4)=0

Now, we will take the common inside the brackets between the start two values and end two values.

We will take x from the start values and we will get it (x²-1), take common -4 from the last two values and we will get it (x²-1). 

x(x²(x²-1) -4(x²-1))=0

Now take common x²-1 from these two values. We will get now x²-4. 

x(x²-1)(x²-4)=0

Now we can see that This is becoming an a²-b² formula which is equal to (a²-b²) = (a+b)(a-b). We can write as.

x(x²-1²)(x²-2²)=0

We can open this formula in (a+b) (a-b) the values will be open in 

x(x+1)(x-1)(x+2)(x-2)=0

Now take these values is equal to zero 

we will get,

x=0 ,

x+1=0 or x=-1,

x-1=0 or x=1,

x+2=0 or x=-2,

x-2=0 or x=2

Here are 5 solutions to this question.

I hope this blog will be very useful to all of you.  If you like this solution comments below.Thankyou

You can also watch this lesson: