THe hypotenuse of a right triangle is 26cm. THe sum of the other two sides is 34 cm. Find the lengths of the other two sides of the triangle.

x^2+(34-x)^2 = 26^2 find x

if a & b are sides, then: a+b=34 a^2 +b^2 = 26^2 solve it for a & b

you can re-write: a=34-b and substitute a in second equation. solve quadratic equation for b & find a

Okay, I did that and then I got 2x^2-68x+1156=676

i got: 2b^2 - 68b+480=0

But I think I did something wrong... because it didn't work out...

The first step, (that dhatraditya wrote) was correct... but something in my process work was wrong

(34-b)^2+b^2=26^2 34^2-68b+2b^2-26^2=0 2b^2-68b+480=0 b^2-34b+240=0

b^2-34b+240 = 0 b = 10, 24

it will give you 10 & 24 ! dhatraditya is right!

Okay, that works, thanks!

you are welcome

Wait, sorry, how do you get from b^2-34b+240 = 0 to b=10, 24 ????

Please answer!!!

do you know how to factor quadratic equations?

yes. I'm pretty sure that's where I'm doing it wrong - I got 2(x-30)(x-4) , but that's not right...

or the quadratic formula. either is okay

I need to factor it for this problem...

no that is not right. 30 * 4 is not equal to 240, it is 120 why did you choose 30 and 4?

Because I didn't multiply right!!!! OOOOOOOPS!

so it should be 2(x-24)(x-10), right?

why are you multiplying the whole thing by 2?

Because I factored 2 out of the equation earlier...

if 2 times x is 0 then x is 0 anyway, your answer is correct.

Okay, thanks again :)

np

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