For right-angle triangle prism, show that x satisfies the equation 3x2 - 14x +16 =0

bumblebee123

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Joined
Jan 3, 2018
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20
Hi,
Please can you show me step by step how to show the answer? I have substituted till I am blue in the face and still don't get the right answer.
I have a question from the quadratic equations section of GCSE. I am told:
The prism is a right-angled triangle. We have established the length of the prism is found by formula L = 1/3( E - 24x)
The surface area, A cm, of the prism is given by formula A = 12x2 + 12Lx
A = 448 cm
E= 98 cm
Substitute these values ( A and E) into the formulae of L and A to show that x satisfies the equation 3x2 - 14x +16 =0

I have inserted L into Formula A but get nothing like I should be getting. Can someone show me how to get there please?

Thank you in advance.
 

tkhunny

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Staff member
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Apr 12, 2005
Messages
9,778
Hi,
Please can you show me step by step how to show the answer? I have substituted till I am blue in the face and still don't get the right answer.
I have a question from the quadratic equations section of GCSE. I am told:
The prism is a right-angled triangle. We have established the length of the prism is found by formula L = 1/3( E - 24x)
The surface area, A cm, of the prism is given by formula A = 12x2 + 12Lx
A = 448 cm
E= 98 cm
Substitute these values ( A and E) into the formulae of L and A to show that x satisfies the equation 3x2 - 14x +16 =0

I have inserted L into Formula A but get nothing like I should be getting. Can someone show me how to get there please?

Thank you in advance.
Better yet, why don't you show us what you get when you substitute? We'll be happy to help you on your way.
 

bumblebee123

New member
Joined
Jan 3, 2018
Messages
20
Sure.

I have substituted A and L formulae with the given values and then substituted L with the formula. I have tried to expand the brackets where the 12lx is concerned but I don't think I am expanding it right.
448 = 12x2 + 12(1/3( 98 - 24x))x

I just don't know how to go any further. I apologise for the highlighted text I don't know where it came from and can't turn it off. Your help is much appreciated.
 

ksdhart2

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Mar 25, 2016
Messages
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What you've done so far is good. That takes care of the easy part of plugging things in. Next comes the slightly more difficult (but not too onerous, hopefully) part of simplifying it down. Let's isolate the terms and take a moment to look only at the right-most term 12(1/3(98 - 24x))x. This is nothing more than a whole bunch of individual terms multiplied together. Perhaps if we rewrite it, getting rid of some extraneous parentheses and being more explicit about inserting multiplication symbols, the next step will become clear:

\(\displaystyle 12 \cdot \dfrac{1}{3} \cdot (98 - 24x) \cdot x\)

The first two terms are literally just numbers, so that part's trivial. Because of the commutative property of multiplication, we can always freely change the order of terms, so let's do that too:

\(\displaystyle 4x \cdot (98 - 24x)\)

At this point, the distributive property/lawhttps://www.mathsisfun.com/definitions/distributive-law.html is your friend. After completing that, the entire right-hand side of the equation will consist of a bunch of terms added together. To finish up, it's simply a matter of collecting like terms and moving things to the other side of the equation, as necessary. Give it another go and see what you get.
 

bumblebee123

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Joined
Jan 3, 2018
Messages
20
Thank you for your help. It explains the last part perfectly.
So I understand what to do to continue:
[FONT=MathJax_Main]4[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]⋅[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]98[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]24[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])[/FONT]







392x - 96x^2

If I put this together with the rest of the equation:

448 = 12x^2 +392x -96x^2

Try to group like terms

448 = -84x^2 + 392x

Try and make it look like what it should (3x2 - 14x +16 =0)

84x^2 - 392x + 448 = 0

I quickly spot that 28 is a factor here so if I divide all coefficients by 28 I get what I am looking for. Yay!

However, I don't understand why I would do that without knowing I have to reach a certain outcome. What is the rule or why would you do this. In questions I have had before I have never done that. I guess if you are multiplying or dividing by a common factor throughout then the equation still becomes balanced?
[FONT=MathJax_Main]4[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]⋅[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]98[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]24[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])
[/FONT]



 
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