GCSE questions

How would you work out:

3.47^2 + 2 x 3.47 x 1.53 + 1.53^2

This is if the first bit of the question said:
Expand and simplify
(x+y)^2

Then it said, hence or otherwise find the value of
3.47^2 + 2 x 3.47 x 1.53 + 1.53^2

By the way this is on the non-calculator paper.

Also how would you simplify fully:

(n^2-1 / n+1) x (2/n-2)

Reply 1

potspansamy

10

The expansion is:
x^2+2xy+y^2
For the second part you may be able to see that it is in the same form as the expansion so if you let x=3.47 and y=1.53 ,
you can put this into the non expanded form
(x+y)^2
this should help you to find the solution.

I am not sure about the second part, (n^2-1 / n+1) x (2/n-2) is that meant to be (n^(2-1) / n+1) x (2/n-2)??

Reply 2

Gaz031

15

*Princess*

How would you work out:

3.47^2 + 2 x 3.47 x 1.53 + 1.53^2

This is if the first bit of the question said:
Expand and simplify
(x+y)^2

Then it said, hence or otherwise find the value of
3.47^2 + 2 x 3.47 x 1.53 + 1.53^2

By the way this is on the non-calculator paper.

Use the identity you wrote in the first part of the question, by letting x and y take appropriate values.

Also how would you simplify fully:

(n^2-1 / n+1) x (2/n-2)

Write n^2-1 as a product of two polynomial factors and simplify by cancelling.

Reply 3

insparato

15

*Princess*

How would you work out:

3.47^2 + 2 x 3.47 x 1.53 + 1.53^2

This is if the first bit of the question said:
Expand and simplify
(x+y)^2

Then it said, hence or otherwise find the value of
3.47^2 + 2 x 3.47 x 1.53 + 1.53^2

By the way this is on the non-calculator paper.

Also how would you simplify fully:

(n^2-1 / n+1) x (2/n-2)

(x+y)² = x² + 2xy + y²

3.47^2 + 2 x 3.47 x 1.53 + 1.53^2

recognise that let x = 3.47 and y = 1.53

thus if (x+y)² = x² + 2xy + y²

then 3.47^2 + 2 x 3.47 x 1.53 + 1.53^2 = (3.47 + 1.53)²

= 5² = 25

(n^2-1 / n+1) x (2/n-2)

(n+1)(n-1)/(n+1) x 2/(n-2) -- difference of two squares

(n+1) x 2/(n-2)

2(n+1)/(n-2)

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