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    (Original post by phat-chewbacca)
    Try dis!
     x= 7 \pm \sqrt 3 \text{ cm} .
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    (Original post by phat-chewbacca)
    Try dis!
    2.20 and 11.8
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    lets try doing non calculator questions only, as that paper is first
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    (Original post by ihatehannah)
    2.20 and 11.8
    Yep . Fuc* Edexcel!
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    Has anyone got an Edexcel account. I really need GCSE Maths 9-1 Sample Assessments (Issue 1).
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    Find the value of  x such that
     \displaystyle x^2-3x+4\sqrt 3 = 0 ,
    giving your values of  x in their exact form.
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    (Original post by phat-chewbacca)
    Yep . Fuc* Edexcel!
    I wonder where you got that question from.....
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    (Original post by ihatehannah)
    I wonder where you got that question from.....
    Solve
     \displaystyle x^4 + 3x^2 -2 =0 .
    ....
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    (Original post by Ano123)
    Solve
     \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!
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    More hard questions pls!
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    (Original post by phat-chewbacca)
    Try dis!
    thats calculator right?


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


    Posted from TSR Mobile
    Yeah boi!
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    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
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    (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!
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    (Original post by phat-chewbacca)
    I'm fine!
    Please explain the answer to the question you posted earlier.
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    (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.
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    (Original post by hamza772000)
    Please explain the answer to the question you posted earlier.
    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!
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    (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?
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    (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 and the other is  \sqrt{144-x^2} (Pythagoras). Then simply solve the equation
     x+\sqrt{144-x^2}=14 .
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    (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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