For what real values of $c$ is $x^2 - 8x - 4x + c + x^2 - 20x + x^2$ the square of a binomial?

tomtom Dec 13, 2023

#1**0 **

The given expression can be simplified by combining like terms:

x2+x2+x2−8x−4x−20x+c=3x2−32x+c

For this expression to be the square of a binomial, it must be of the form (ax+b)2. Expanding this form gives us:

(ax+b)2=a2x2+2abx+b2

Comparing the coefficients of the term in x2, the term in x, and the constant term in both expressions, we get:

a2=3

2ab=−32

b2=c

Solving the first equation, we have a=±3.

From the second equation, we can substitute a=3 or a=−3 and solve for b:

For a=3: 23b=−32, so b=−316=−3163.

For a=−3: −23b=−32, so b=316=3163.

Therefore, the expression can be the square of a binomial with two different sets of coefficients:

(x3−3163)2=3x2−32x+128/3, which occurs when c=128/3.

((−x)3+3163)2=3x2−32x+128/3, which occurs when c=128/3.

However, it's important to note that the square of a binomial cannot have a negative sign in front of the quadratic term. Therefore, the only valid solution for c is:

c= 128/3

Boseo Dec 13, 2023