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Please help me with this problem..!

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!

Reply 1

liv20

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 :smile:

Reply 2

Aegyst

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 :smile:

Yes, I also got the equation. But I couldn't solve it. Could you please explain?
Thanks, by the way...

Reply 3

wolfmoon88

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

Posted from TSR Mobile

Reply 4

RogerOxon

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 (

x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}

). One root will be negative, so should be discounted. Please post your working if you have any issues.

(edited 6 years ago)

Reply 5

Aegyst

Original post by wolfmoon88

expand (x+1)^2 first

Posted from TSR Mobile

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?

Reply 6

B_9710

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).