Question

If(a²-1) x²+(a-1)x+a²-4a+3=0 is an identity in x, then find the value of a​

Answer 1

Answer:

Step-by-step explanation:

$(a^2-1)x^2+(a-1)x+a^2-4a+3=0\\\\Calculate\ and\ identify\ the\ polynomials\\\\\Longleftrightarrow\ a^2x^2-x^2+ax-x+a^2-4a+3=0\\\\\Longleftrightarrow\ a^2x^2+ax+a^2-4a+3=x^2+x+0\\\\\Longleftrightarrow\ \left\{\begin{array}{ccc}a^2&=&1\\a&=&1\\a^2-4a+3&=&0\\\end{array} \right.\\\\\Longleftrightarrow\ \left\{\begin{array}{ccc}(a-1)(a+1)&=&0\\a-1&=&0\\(a-1)(a-3)&=&0\\\end{array} \right.\\\\\\We\ must\ exclude\ a=-1\ and\ a=3\ (not\ solution)\\\Longrightarrow\ a=1$