C-05-Q102mediumsingle_mcqComparing x2−1=0x^2 - 1 = 0x2−1=0 with ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0, we get—aa=1,b=0,c=−1a = 1, b = 0, c = -1a=1,b=0,c=−1ba=1,b=1,cব্যাখ্যাWriting x2−1=0x^2-1=0x2−1=0 as 1⋅x2+0⋅x+(−1)=01\cdot x^2 + 0\cdot x + (-1)=01⋅x and matching with gives , , .