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# Please help me with this problem..! watch

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1. Anyone, please help me with this question:

The shortest side AB of a right-angled triangle is x cm long.
The side BC is 1 cm longer than AB, and the hypothenuse AC is 29 cm long.
Form an equation for x and solve it to find the length of each sides!

I was able to form the equation, but couldn't find the value of x.

Thanks!
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2. so AB = x and BC = x+1 and AC = 29
so (x^2) + (x+1)^2 = 29^2
form a quadratic and solve from there
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3. (Original post by leviaxx)
so AB = x and BC = x+1 and AC = 29
so (x^2) + (x+1)^2 = 29^2
form a quadratic and solve from there
Yes, I also got the equation. But I couldn't solve it. Could you please explain?
Thanks, by the way...
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4. (Original post by Aegyst)
Yes, I also got the equation. But I couldn't solve it. Could you please explain?
Thanks, by the way...
expand (x+1)^2 first

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5. (Original post by Aegyst)
Yes, I also got the equation. But I couldn't solve it. Could you please explain?
Re-arrange to get a a quadratic, which you can solve with the formula (). One root will be negative, so should be discounted. Please post your working if you have any issues.
6. (Original post by wolfmoon88)
expand (x+1)^2 first

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Is this right? :

x²+(x+1)²-841=0
x²+x²+2x+1-840=0
2x²+2x-840=0

Could you explain the next step, please?
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7. (Original post by Aegyst)
Is this right? :

x²+(x+1)²-841=0
x²+x²+2x+1-840=0
2x²+2x-840=0

Could you explain the next step, please?
Divide the equation by 2 then it does actually factorise and then just factorise and solve. You'll get one negative answer which you can forget about because ofc length can't be negative (context of the question is importantl for future questions).
8. (Original post by B_9710)
Divide the equation by 2 then it does actually factorise and then just factorise and solve. You'll get one negative answer which you can forget about because ofc length can't be negative (context of the question is importantl for future questions).
Okay, I got it.
Thanks for your help!
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