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Can someone help me with this question please?

Hi all, I need help with this question. Been trying for ages with no luck.

Using Pythagoras' Theorem, A = 4(x + 2), B = (x + 3), C = 25. Find X.

help pls
Reply 1
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Zacken
22
Original post by OwlnMcgee
Hi all, I need help with this question. Been trying for ages with no luck.

Using Pythagoras' Theorem, A = 4(x + 2), B = (x + 3), C = 25. Find X.

help pls


You know a^2 + b^2 = c^2. You also know how to solve a quadratic.
Reply 2
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OwlnMcgee
OP
Original post by Zacken
You know a^2 + b^2 = c^2. You also know how to solve a quadratic.


Yep. But I can't seem to actually solve the question.
Reply 3
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Zacken
22
Original post by OwlnMcgee
Yep. But I can't seem to actually solve the question.


What stage have you reached? You know that (4(x+2))2+(x+3)2=252(4(x+2))^2 + (x+3)^2 = 25^2
Reply 4
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OwlnMcgee
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Original post by Zacken
What stage have you reached? You know that (4(x+2))2+(x+3)2=252(4(x+2))^2 + (x+3)^2 = 25^2


I don't know if this is right, but:

4(x^2 + 2x + 4) + (x^2 + 3x + 3x + 9) = 25^2

Another version that I ended up with was 5x^2 + 22x = 600.

I've got really confused and ended up with many different outcomes, none actually being "final".
Reply 5
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Zacken
22
Original post by OwlnMcgee
I don't know if this is right, but:

4(x^2 + 2x + 4) + (x^2 + 3x + 3x + 9) = 25^2

Another version that I ended up with was 5x^2 + 22x = 600.

I've got really confused and ended up with many different outcomes, none actually being "final".


(4(x+2))2=4(x+2)⋅4(x+2)=16(x2+4x+4)(4(x+2))^2 = 4(x+2) \cdot 4(x+2) = 16(x^2 + 4x + 4)

The second bracket is correct. So now simplify and solve.
Reply 6
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OwlnMcgee
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Original post by Zacken
(4(x+2))2=4(x+2)⋅4(x+2)=16(x2+4x+4)(4(x+2))^2 = 4(x+2) \cdot 4(x+2) = 16(x^2 + 4x + 4)

The second bracket is correct. So now simplify and solve.



Okay, thanks. I got to 17x^2 + 70x = 552. What do I do now?
Reply 7
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Zacken
22
Original post by OwlnMcgee
Okay, thanks. I got to 17x^2 + 70x = 552. What do I do now?


Assuming that's correct, it's the same thing as 17x2+70x−552=017x^2 + 70x - 552 = 0 which is a quadratic. Now... quadratic formula.
Reply 8
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OwlnMcgee
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Original post by Zacken
Assuming that's correct, it's the same thing as 17x2+70x−552=017x^2 + 70x - 552 = 0 which is a quadratic. Now... quadratic formula.



Yeah, I got to that. Can you please help me with the quadratic equation? I'm not too great at complex ones.
Reply 9
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Zacken
22
Original post by OwlnMcgee
Yeah, I got to that. Can you please help me with the quadratic equation? I'm not too great at complex ones.


Have you heard of the quadratic formula? Can you apply it in this scenario with
a = 17, b=70, c = -552?
Reply 10
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OwlnMcgee
OP
Original post by Zacken
Have you heard of the quadratic formula? Can you apply it in this scenario with
a = 17, b=70, c = -552?


Yeah, I'm just looking at it now, haha. I'll give it a shot. (only learned quadratic equations like, three days ago so I'm not experienced with them yet)
Reply 11
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Zacken
22
Original post by OwlnMcgee
Yeah, I'm just looking at it now, haha. I'll give it a shot. (only learned quadratic equations like, three days ago so I'm not experienced with them yet)


Not a problem, don't worry. What do you get? The quadratic formula says that if you had a quadratic equation ax2+bx+c=0ax^2 + bx + c = 0, then your solutions are

x=b±b2−4ac2a\displaystyle x = \frac{b \pm \sqrt{b^2 - 4ac}}{2a}
Reply 12
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OwlnMcgee
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Uhhh, I've tried multiple different ways of using those numbers in the quadratic formula, but still don't get the right answer.

Worst thing is, I get TWO wrong answers instead of one! :frown:
@Zacken, you're a good kid. Stop letting the powers of community assistant get inside your head. I don't know the full story of the traumatic events that took place at your school leading you to drop out. However, @physicsmaths has told me a bit about the bullies and I couldn't imagine how hard it must've been for you. Also, please don't take some of my immature posts too seriously, it's all banter. :tongue:

Take care bud!
Reply 14
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Zacken
22
Original post by OwlnMcgee
Uhhh, I've tried multiple different ways of using those numbers in the quadratic formula, but still don't get the right answer.

Worst thing is, I get TWO wrong answers instead of one! :frown:


Do you mind posting your working so I can see where you went wrong?
Reply 15
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OwlnMcgee
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Original post by Zacken
Do you mind posting your working so I can see where you went wrong?


Oh. I just tried it again and got it right.

What was happening was first of all I got the operators wrong, and then it turns out I had to clear my calculator every time.

Thank you all though! :smile:

note: the answer was 4 cm
(edited 8 years ago)
Reply 16
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Zacken
22
Original post by OwlnMcgee
Oh. I just tried it again and got it right.

What was happening was first of all I got the operators wrong, and then it turns out I had to clear my calculator every time.

Thank you all though! :smile:

note: the answer was 4 cm


Great! Glad you got it. :smile:
Original post by OwlnMcgee
Oh. I just tried it again and got it right.

What was happening was first of all I got the operators wrong, and then it turns out I had to clear my calculator every time.

Thank you all though! :smile:

note: the answer was 4 cm


Correct! Do you understand what wast done and why we arrived at the answer?
Reply 18
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OwlnMcgee
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Yeah, I think so. Thanks again! :smile:

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