Question

If $m-nx+28x^2+12x^3+9x^4$ is a perfect square, then find the values of m and n.

• A. $n=-16$ and $m=16$
• B. $n=16$ and $m=16$
• C. $n=-16$ and $m=-16.$
• D. None of these

Answer 1

Answer: (A)

$n=-16$ and $m=16$

It is given that the polynomial $m-nx+28x^2+12x^3+9x^4$ is a perfect square, we must equate it to the square of general form of equation that is $(a{x}^2+bx+c{)}^2$ as shown below:

$9x^4+12x^3+28x^2-nx+m={(a{x}^2+bx+c)}^2\Rightarrow 9x^4+12x^3+28x^2-nx+m={\left(a{x}^2\right)}^2+{\left(bx\right)}^2+{\left(c\right)}^2+(2\times a{x}^2\times bx)+(2\times bx\times c)+(2\times c\times a{x}^2)(\because (a+b+c{)}^2=a^2+b^2+c^2+2ab+2bc+2ca)\Rightarrow 9x^4+12x^3+28x^2-nx+m=a^2{x}^4+b^2{x}^2+c^2+2ab{x}^3+2bcx+2ac{x}^2$

Now, comparing the coefficients, we get:

$a^2=9,b^2+2ac=28,c^2=m,2ab=12,2bc=-n$

$a^2=9\Rightarrow a=3$

$2ab=12\Rightarrow 2\times 3\times b=12\Rightarrow 6b=12\Rightarrow b=2$

$b^2+2ac=28\Rightarrow 2^2+(2\times 3c)=28\Rightarrow 4+6c=28\Rightarrow 6c=28-4\Rightarrow 6c=24\Rightarrow c=4$

$c^2=m\Rightarrow m=4^2\Rightarrow m=16$

$2bc=-n\Rightarrow n=-2bc\Rightarrow n=-2\times 2\times 4\Rightarrow n=-16$

Hence, $m=16$ and $n=-16$.