If $a<c<b$ the the roots of the equations $(a-b)^2x^2+2(a+b-2c)x+1=0$ are ?

a < c < b

$(a-b)^{2} x^{2}+2(a+b-2 c) x+1=0$

নিশ্চায়ক, $D=b^{2}-4 a c$

$\begin{aligned} & =\left\{2(a+b-2 c)^{2}-4(a-b)^{2} \cdot 1\right. \\ & \left.=4\left\{(a+b)^{2}-2 \cdot 2 c(a+b)+4 c^{2}\right\}^{2}\right\}-4\left(a^{2}-2 a b+b^{2}\right) \\ & =4\left(a^{2}+2 a b+b^{2}-4 a c-4 b c+4 c^{2}\right)-4 a^{2}+8 a b-4 b^{2} \\ & =4 a^{2}+8 a b+4 b^{2}-16 a c-16 b c+16 c^{2}-4 a^{2}+8 a b -4 b^{2}\\ & =16 a b-16 a c-16 b c+16 c^{2} \\ & =16\left(a b-a c-b c+c^{2}\right) \\ & =16\left\{a(b-c)-c\left(b-c\right\}\right. \\ \therefore D & =16(b-c)(a-c) \end{aligned}$

$\ \therefore$ a < c সুতরাং a - c এর মান হবে Negative অর্থাৎ D এর মান তাহলে Negative আসবে।

আমরা জানি, D <0 হলে মূলদ্বয় হবে কাল্পনিক অর্থাৎ imaginary।