### Question 5

### 6 comments

###### Goodness Barak

1 year ago

@Oluwasegun Solaja that will be a nice one.

Sure I'll let ppl know abt the platform I'm doing that already.

###### Oluwasegun Solaja

1 year ago

@Goodness Barak Don't you think you can support this forum to grow by getting people to know more about this platform and solve be solving Mathematics problems?

We can as well work on having Naitalk Video on Mathematics to solve problems for students. We can all correct deficiencies in the education sector in Nigeria.

###### Goodness Barak

1 year ago

Let x^(1/5) = y

Substitute into the equation we have

y²-5y+6=0

Solving using factorisations

y²-2y-3y+6=0

y(y-2)-3(y-2)=0

(y-3)(y-2)=0

y-3=0 or y-2=0

y=3 or y=2

Since, y=x^(1/5)

Hence,

When y=3 we have

3=x^(1/5)

Multiply powers of both sides by 5

3^5=x

x=243

Also when y=2 we have

x^(1/5)=2

Multiply the powers of both sides by 5

x=2^5

x=32

Therefore,

x=243 or 32