Question

If $f\left(x\right)=4x^4+12x^3+c{x}^2+6x+d$ is a perfect square, then

• A. $c=5$
• B. $c=13$
• C. $d=1$
• D. Minimum value of $f\left(x\right)$ occurs at $x=-\frac{1}{2}$

Answer 1

Answer: (B), (C), (D)

Minimum value of $f\left(x\right)$ occurs at $x=-\frac{1}{2}$

$f\left(x\right)=4x^4+12x^3+c{x}^2+6x+d$
If a bi-quadratic polynomial is a perfect square, then it must be in the form of
$(a{x}^2+px+q{)}^2$
$=a^2{x}^4+2ap{x}^3+(p^2+2aq)x^2+2pqx+q^2$
Comparing the coefficient, we get
$a^2=4\Rightarrow a=\pm 2$
$2ap=12\Rightarrow p=\pm 3$
$2pq=6\Rightarrow q=\pm 1$
$\Rightarrow a,p,q$ have same sign.

Now, $d=q^2\Rightarrow d=1$
$c=p^2+2aq$
$=9+2(\pm 2)(\pm 1)$
$=9+4$
$=13$

$f\left(x\right)=(2x^2+3x+1{)}^2$
$=4x^4+12x^3+13x^2+6x+1$
The minimum value of $f\left(x\right)$ is $0$
$(2x^2+3x+1{)}^2=0$
$\Rightarrow 2x^2+3x+1=0$
$\Rightarrow (2x+1)(x+1)=0$
$\Rightarrow x=-\frac{1}{2}\text{or}-1$