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9x²-9(a+b)x+(2a²+5ab+2b²)=0...,.quardic equation...plz solve

Answer» There are two values of x..X=(2a+b)/3 or X=(a+2b)/3... Here the steps..D=b2-4ac ...now...put the values as b=-9(a+b) or a=9 and c=(2a2+5ab+2b2)..After that u will get D=(3a-3b) whole square..now u can find its zeros..by alpha=√D-b/2a..or beta =-√D-b/2a..


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