Let, ABC is the right angle triangle
Here, Hypotenuse AB = (2x - 6)
One arm, AC = x And another arm, BC = (x + 2) ∴According to Pethagorean Theorem, (2x-6)2=(x+2)2+x2
=(2x)2-2×2x×6+62=x2+2×x×2+22+x2
=4x2-24x+36=2x2+4x+4 =2x2-28x+32=0 =x2-14x+16=0
Now, comparing, x2-14x+16=0 with ax2+bx+c=0, We get, x=-b±b2-4ac2a=-(-14)±(-14)2-4×1×162×1=14±196-642=14±1322=14±4×332
=14±2332=2(7±33)2=7±33 ∴x=(7+33) or (7-33)
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