Question

If 81x⁴-72x³+px²-8x+1 is a perfect square the find the value of p

Answer 1

$\textbf{Given:}$

$\text{ 81x^4-72x^3+p\,x^2-8x+1 is a perfect square}$

$\textbf{To find:}$

$\text{The value of 'p'}$

$\textbf{Solution:}$

$\text{Since 81x^4-72x^3+p\,x^2-8x+1 is a perfect square,}$

$\text{it can be written as}$

$81x^4-72x^3+p\,x^2-8x+1=(9\,x^2+m\,x+1)^2$

$81x^4-72x^3+p\,x^2-8x+1=(9\,x^2+m\,x+1)(9\,x^2+m\,x+1)$

$\text{Equating coefficient of x on both sides, we get}$

$-8=m+m$

$2m=-8$

$m=\dfrac{-8}{2}$

$\implies\bf\,m=-4$

$\text{Equating coefficient of x^2 on both sides, we get}$

$p=9+9+m^2$

$p=9+9+(-4)^2$

$p=9+9+16$

$\implies\boxed{\bf\,p=34}$

$\therefore\textbf{The value of is 34}$

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