Find all real solutions of the equation:

[tex](6-2x)(3-2x)x = 40[/tex]

A. Yes. Because [tex]\frac{3}{2}[/tex] is a root, you can cut squares with sides of [tex]\frac{3}{2}[/tex] in. to make the box.

B. No. This equation has no real solutions.

C. No. The only real solution is [tex]x=4[/tex]. It is not possible to cut squares of this size.



Answer :

To solve the equation [tex]\( (6 - 2x)(3 - 2x)x = 40 \)[/tex], we proceed as follows:

1. Distribute and Form the Polynomial Equation: The given equation is [tex]\((6 - 2x)(3 - 2x)x = 40\)[/tex].
Let's first consider the product [tex]\((6 - 2x)(3 - 2x)\)[/tex]:

[tex]\[ (6 - 2x)(3 - 2x) = 6 \cdot 3 - 6 \cdot 2x - 2x \cdot 3 + (-2x)(-2x) \][/tex]
[tex]\[ = 18 - 12x - 6x + 4x^2 \][/tex]
[tex]\[ = 4x^2 - 18x + 18 \][/tex]

2. Substitute into the Original Equation: Now, insert this into the original equation:
[tex]\[ (4x^2 - 18x + 18)x = 40 \][/tex]
[tex]\[ 4x^3 - 18x^2 + 18x = 40 \][/tex]

3. Rearrange to Form a Standard Polynomial Equation:
[tex]\[ 4x^3 - 18x^2 + 18x - 40 = 0 \][/tex]

4. Solve the Polynomial Equation: Next, we need to find the real roots of the polynomial [tex]\(4x^3 - 18x^2 + 18x - 40 = 0\)[/tex]. By solving this cubic equation, we find the real solution:

[tex]\[ x = 4 \][/tex]

Thus, the only real solution to the equation [tex]\( (6 - 2x)(3 - 2x)x = 40 \)[/tex] is [tex]\( x = 4 \)[/tex].

Regarding the context of the question:
- The real solution [tex]\( x = 4 \)[/tex] does not indicate that there is a suitable size to cut the squares (e.g., for a box), as the solution [tex]\( x = 4 \)[/tex] may not fit the practical constraints of the problem context such as dimensions or other given constraints.

Therefore, the correct response to the given context is:

No. The only real solution is [tex]\( x = 4 \)[/tex]. It is not possible to cut squares of this size.