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Math Help - Quadratics and simultaneous equations

  1. #1
    BoF
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    Quadratics and simultaneous equations

    Got two questions here that are really puzzling me.


    1. The perpendicular height of a triangle is 3cm less than the lenght of its base. The area of the triangle is 27cm (squared). Form a quadratic equation to represent the information and solve to find the dimensions of the triangle.

    next one is

    2. The area of a rectangle is 21 cm (squared) and its perimeter is 20 cm.
    The length of the rectangle is x cm and its width is y cm.

    a Write equations for the area and perimeter in terms of X and Y

    b Use substitution in order to form a quadratic in one variable

    c Solve to find the midensions of the rectangle.

    Ive attempted both questions but am unable to come to the correct answer
    If someone could please answer these questions aswell as show me how to do it, it would save me alot of headaches

    Thank you
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  2. #2
    Super Member Matt Westwood's Avatar
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    Which are the specific bits that are puzzling you? I'll start you off:

    a) Area of triangle is half base times height.

    So let the base be b and the height be h, and the area be A.

    Then A = \frac {bh}2

    But then h = b-3 and A=27.

    You can build your quadratic from there by substituting b-3for h and 27 for A in the equation for the area.

    b) The area of a rectangle is height times width. You've been given the variables for height and width, they're x and y. So area = x times y.

    As for the perimeter, that's the line that goes round the outside. There's two lengths and two widths. Add those up together and that's the perimeter.

    Take it from there.
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  3. #3
    BoF
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    Thx for the quick answer but ironically the things you said are the thing i do understand, its the next parts which i cant do so

    1) form a quadratic equation to represent the information and solve to find the dimensions of the triangle

    and

    2) Use substitution in order to form a quadratic in one variable + solve to find dimensions of the rectangle

    these are the things im not sure how to do.
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  4. #4
    BoF
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    lol been using elimination on question 2 rather than substitution lol

    now to find out what substitution is...
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  5. #5
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by BoF View Post
    Thx for the quick answer but ironically the things you said are the thing i do understand, its the next parts which i cant do so

    1) form a quadratic equation to represent the information and solve to find the dimensions of the triangle

    and

    2) Use substitution in order to form a quadratic in one variable + solve to find dimensions of the rectangle

    these are the things im not sure how to do.
    1) Once you've done what I suggested, you'll see a quadratic equation staring you in the face. So you just solve the quadratic the way you've been taught to solve quadratic equations.

    2) Do you understand what the word "substitution" means? Imagine you were watching a game of football. During the course of the game, one of the members of one of the team chips his nail-varnish or his perm gets mussed. Obviously he can't continue playing, so he has to come off the field to get his make-up fixed. But the game (apparently) has to carry on without him, and it isn't fair to make a team play with one man too few. So someone else takes his place. That is, he is substituted. It's exactly the same with equations. If you know that something is equal to something else, then you can substitute one for the other.
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  6. #6
    BoF
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    well to be honest i do understand the word substitution but maby were not talking bout same thing.

    i will show you what i am doing and where im going wrong

    A = x * Y = 21 (squared) (1)
    P = 2x + 2y = 20 (2)

    (1) x2

    2x * 2y = 42 (squared) (3)
    + 2x + 2y = 20 (2)
    --------------------------------
    4x 4y = 62

    X = 15.5
    y= 15.5

    This is pretty much the point i get to before im extremly lost, as for 1) im not sure if its the way you worded it or just me not fully understanding how to put it into a quadratic but im not sure what to do with it.
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  7. #7
    Super Member Matt Westwood's Avatar
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    Right, now we're getting somewhere. It's always best when asking a question on here to explain what you've done so far, it gives us poor workers a starting point ...

    A = 21 = xy

    So y = \frac {21} x

    Now substitute for y in the other equation:

    P = 20 = 2x + 2y = 2x + 2 \left({\frac {21} x}\right)


    So you have the equation 2x + 2\left({\frac {21} x}\right) = 20

    Just twiddle the equation into something that looks like ax^2 + bx + c = 0.

    Start by multiplying the whole thing by x (both sides, mind) to clear the fraction and then it's just gathering the terms.

    Mind, by looking at it, you can probably work out what you expect the answer's going to be by thinking of two numbers that multiplied together make 21 and added together make 10 ... as you do more of these things it will all start making sense.
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  8. #8
    BoF
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    K thx alot ^^
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  9. #9
    Super Member Matt Westwood's Avatar
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    If you're done, feel free to set the thread as "Solved" (somewhere in the tools menu above the thread) and click a "thanks" button to any posting which helped you in particular. It's not compulsory, so feel free not to.
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