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1)x^2-2ax+a^2-b^2=0
=> x^2-2ax+a^2-b^2=0
=>x^2-2ax+a^2-b^2=0
=>(x-a)^2-b^2=0
now use formula
a^2-b^2=(a+b)(a-b)
so, {x-a-b}{x-a+b}=0
x=a+b, a-b
2)in the same way
(x^2-4ax+4a^2)-b^2=0
(x-2a)^2-b^2=0
{x-2a-b}{x-2a+b}=0
x=2a+b,2a-b
now I think you understand concept how this proceed. so I directly goes to answer
3) {2x-a-b}{2x-a+b}=0
x=(a+b)/2, (a-b)/2
4)x=(a^2+b^2)/2, (a^2-b^2)/2
=> x^2-2ax+a^2-b^2=0
=>x^2-2ax+a^2-b^2=0
=>(x-a)^2-b^2=0
now use formula
a^2-b^2=(a+b)(a-b)
so, {x-a-b}{x-a+b}=0
x=a+b, a-b
2)in the same way
(x^2-4ax+4a^2)-b^2=0
(x-2a)^2-b^2=0
{x-2a-b}{x-2a+b}=0
x=2a+b,2a-b
now I think you understand concept how this proceed. so I directly goes to answer
3) {2x-a-b}{2x-a+b}=0
x=(a+b)/2, (a-b)/2
4)x=(a^2+b^2)/2, (a^2-b^2)/2
faridha if you like please take brainliest
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