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Edexcel GCSE maths, give each other questions to do.

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Reply 40
Original post by phat-chewbacca
Try dis!


x=7±3 cm x= 7 \pm \sqrt 3 \text{ cm} .
Original post by phat-chewbacca
Try dis!


2.20 and 11.8
lets try doing non calculator questions only, as that paper is first
Original post by ihatehannah
2.20 and 11.8
Yep :smile:. Fuc* Edexcel!
Has anyone got an Edexcel account. I really need GCSE Maths 9-1 Sample Assessments (Issue 1).
Reply 45
Find the value of x x such that
x23x+43=0 \displaystyle x^2-3x+4\sqrt 3 = 0 ,
giving your values of x x in their exact form.
(edited 7 years ago)
Original post by phat-chewbacca
Yep :smile:. Fuc* Edexcel!


I wonder where you got that question from.....
Reply 47
Original post by ihatehannah
I wonder where you got that question from.....


Solve
x4+3x22=0 \displaystyle x^4 + 3x^2 -2 =0 .
....
Original post by Ano123
Solve
x4+3x22=0 \displaystyle x^4 + 3x^2 -2 =0 .
....

Here is my workings!

Solve x^4 +3x^2 -2 =0

Let t = x^2

Therefore t^2 +3t -2 =0

t = (-3 +√17) / 2 or t = (-3-√17) / 2

Therefore x^2 = (-3 +√17) / 2 or x^2 = (-3-√17) / 2

x = √(-3 +√17) / 2 or x = √(-3-√17) / 2 Discard this value.


x = 0.749368275


A* here I come!
More hard questions pls!
Reply 50
Original post by phat-chewbacca
Try dis!


thats calculator right?


Posted from TSR Mobile
Original post by ksj.11
thats calculator right?


Posted from TSR Mobile
Yeah boi!
Judging by the new style of questions…

If 1+1=2,

What is the Force on the Sun when hydrostatic pressure of 1023222Pa is exerted on a surface area of 2m^2?

Give your answer to 3 s.f. when 2+2=4
Original post by Bigbosshead
Judging by the new style of questions…

If 1+1=2,

What is the Force on the Sun when hydrostatic pressure of 1023222Pa is exerted on a surface area of 2m^2?

Give your answer to 3 s.f. when 2+2=4
I'm fine! :u:
Original post by phat-chewbacca
I'm fine! :u:
Please explain the answer to the question you posted earlier.:smile:
Reply 55
Original post by phat-chewbacca
Here is my workings!

Solve x^4 +3x^2 -2 =0

Let t = x^2

Therefore t^2 +3t -2 =0

t = (-3 +√17) / 2 or t = (-3-√17) / 2

Therefore x^2 = (-3 +√17) / 2 or x^2 = (-3-√17) / 2

x = √(-3 +√17) / 2 or x = √(-3-√17) / 2 Discard this value.


x = 0.749368275


A* here I come!


You've only put one answer down? (Try and stick to exact values as well).
It may be useful to explain why you have rejected 2 answers.
(edited 7 years ago)
Original post by hamza772000
Please explain the answer to the question you posted earlier.:smile:
Well we were told that the perimeter of rectangle is 28cm. So if we denote the length of the rectangle as y, then the perimeter would be the whole distance around the rectangle. Which would be 2x + 2y.

We now that the perimeter is 2x +2y, therefore we can equate this to 28 as we were told that the perimeter of the rectangle is 28cm.

Equation 1:

2x + 2y =28

Dividing through by two. Therefore x + y = 14. Then we can rearrange the equation making y the subject. Which gives us, y = 14 - x


Another fact the we now is that the width is xcm and the length of the diagonal is 12 cm. As we have let the length of the rectangle equal to y, we can now you Pythagoras which gives us x^2 + y^2 = (12)^2. Which simplifies to x^2 +y^2 =144

We now have two equations.

Equation 1: x + y = 14 and Equation 2: x^2 + y^2 =144. I would now like you to solve these simultaneous equations. Hope this helped!
(edited 7 years ago)
Original post by phat-chewbacca
Well we were told that the perimeter of rectangle is 28cm. So if we denote the length of the rectangle as y, then the perimeter would be the whole distance around the rectangle. Which would be 2x + 2y.

We now that the perimeter is 2x +2y, therefore we can equate this to 28 as we were told that the perimeter of the rectangle is 28cm.

Equation 1:

2x + 2y =28

Dividing through by two. Therefore x + y = 14. Then we can rearrange the equation making y the subject. Which gives us, y = 14 - x


Another fact the we now is that the width is 8cm and the length of the diagonal is 12 cm. As we have let the length of the rectangle equal to y, we can now you Pythagoras which gives us x^2 + y^2 = (12)^2. Which simplifies to x^2 +y^2 =144

We now have two equations.

Equation 1: x + y = 14 and Equation 2: x^2 + y^2 =144. I would now like you to solve these simultaneous equations. Hope this helped!

Quick question: how would you know that the width is 8?
Reply 58
Original post by phat-chewbacca
Well we were told that the perimeter of rectangle is 28cm. So if we denote the length of the rectangle as y, then the perimeter would be the whole distance around the rectangle. Which would be 2x + 2y.

We now that the perimeter is 2x +2y, therefore we can equate this to 28 as we were told that the perimeter of the rectangle is 28cm.

Equation 1:

2x + 2y =28

Dividing through by two. Therefore x + y = 14. Then we can rearrange the equation making y the subject. Which gives us, y = 14 - x


Another fact the we now is that the width is 8cm and the length of the diagonal is 12 cm. As we have let the length of the rectangle equal to y, we can now you Pythagoras which gives us x^2 + y^2 = (12)^2. Which simplifies to x^2 +y^2 =144

We now have two equations.

Equation 1: x + y = 14 and Equation 2: x^2 + y^2 =144. I would now like you to solve these simultaneous equations. Hope this helped!


It is not necessary to solve simultaneous equations for this question.
Simply one of the side lengths is x x and the other is 144x2 \sqrt{144-x^2} (Pythagoras). Then simply solve the equation
x+144x2=14 x+\sqrt{144-x^2}=14 .
Original post by hamza772000
Quick question: how would you know that the width is 8?
Sorry my fault the which is xcm as they stated that in the question. I must have put 8cm by accident. I have edited my written explanation and changed it to xcm. Sorry about that!

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