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,c=−1a = 1, b = 1, c = -1a=1,b=1,c=−1ca=1,b=−1,c=0a = 1, b = -1, c = 0a=1,b=−1,c=0da=−1,b=0,c=1a = -1, b = 0, c = 1a=−1,b=0,c=1ব্যাখ্যাWriting x2−1=0x^2-1=0x2−1=0 as 1⋅x2+0⋅x+(−1)=01\cdot x^2 + 0\cdot x + (-1)=01⋅x2+0⋅x+(−1)=0 and matching with ax2+bx+c=0ax^2+bx+c=0ax2+bx+c=0 gives a=1a=1a=1, b=0b=0b=0, c=−1c=-1c=−1.