Question

find the roots of x2-4ax+4a2-b2=0 if they exit by method of competing the square

Answer 1

this ans may help you

Answer 2

Answer:

Step-by-step explanation:

X^2-4ax+4a^2-b^2=0

x^2-2*2a*x+(2a)^2-(b)^2=0

(x-2a)^2-(b)^2=0

(x-2a-b)(x-2a+b)=0

x-2a-b=0 x-2a+b=0

x=b+2a x=2a-b

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