Question

What is wrong here ?

If t³ = t ( If t = 0 , 1 and -1 ) . The solution satisfies the equation . But if we divide t³= t By ( t ) on both sides. ( 0 ) Didn't Satisfies the equation .

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Answer 1

Answer:

When you do the step

t (t+1) (t-1) = 0

the next step will be

(t+1)(t-1) = 0. ..(1)

It is so because the t is taken to the right side and 0/t gives you zero only.

Now from 1 we get that , t= -1 or 1.

If you put the value of t as -1 or 1 in the formula t³=t then the equation will be true.

Answer 2

Reason

Because division removed a factor, it doesn't have $t=0$ as a root. To be specific, the division is used to reject a solution while the product is used to add a solution.

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Example 1.

We need a different example to explain this. Let's take an example, the cube roots of unity. Say, we need the imaginary solutions only.

Let $x$ be the cube root of unity.

By its definition of cube root, the required equation is $x^3=1$.

Given, $x^3=1$

$\rightarrow x^3-1=0$

$\rightarrow (x-1)(x^2+x+1)=0$

We know the first factor results to $x=1$. To reject this, we divide by $x-1$ to remove the factor.

$\rightarrow x^2+x+1=0$

This leads to two imaginary solutions. So, we removed a real solution by division.

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Example 2.

Let's take a similar example but in the opposite way. Say, we need to know about the property of $x^2+x+1=0$.

We multiply $x-1$ to add a factor.

$\rightarrow (x-1)(x^2+x+1)=0$

$\rightarrow x^3-1=0$

$\rightarrow x^3=1$

So, we observe that the solutions of $x^2+x+1=0$ are the cube roots of unity. This happened because we added a solution from the multiplication.

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This is the required answer.