To analyze the given quadratic equation [tex]\((x + 15)(x) = 100\)[/tex] and determine the reasonableness of the solutions [tex]\(x = 5\)[/tex] and [tex]\(x = -20\)[/tex], follow these steps:
1. Expand the equation:
Start with the given equation:
[tex]\[ (x + 15)(x) = 100 \][/tex]
Distribute [tex]\(x\)[/tex] through the term [tex]\((x + 15)\)[/tex]:
[tex]\[ x \cdot x + x \cdot 15 = 100 \][/tex]
This simplifies to:
[tex]\[ x^2 + 15x = 100 \][/tex]
2. Form the standard quadratic equation:
Rewrite the equation in standard quadratic form [tex]\(ax^2 + bx + c = 0\)[/tex]:
[tex]\[ x^2 + 15x - 100 = 0 \][/tex]
3. Solve the quadratic equation:
Use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 15\)[/tex], and [tex]\(c = -100\)[/tex]:
First, compute the discriminant:
[tex]\[ \Delta = b^2 - 4ac = 15^2 - 4 \cdot 1 \cdot (-100) = 225 + 400 = 625 \][/tex]
Next, find the roots:
[tex]\[ x = \frac{-15 \pm \sqrt{625}}{2 \cdot 1} = \frac{-15 \pm 25}{2} \][/tex]
This gives two solutions:
[tex]\[ x = \frac{-15 + 25}{2} = \frac{10}{2} = 5 \][/tex]
and
[tex]\[ x = \frac{-15 - 25}{2} = \frac{-40}{2} = -20 \][/tex]
4. Evaluate the reasonableness of the solutions:
The solutions to [tex]\( (x + 15)(x) = 100 \)[/tex] are [tex]\( x = 5 \)[/tex] and [tex]\( x = -20 \)[/tex].
- [tex]\(x = 5\)[/tex]: This is a positive solution and is reasonable because a width of 5 units is valid for a picture frame.
- [tex]\(x = -20\)[/tex]: This is a negative solution and is not reasonable because a width of [tex]\(-20\)[/tex] units is physically impossible for a picture frame.
Therefore, the correct statement is:
The solution [tex]\(x = 5\)[/tex] should be kept, but [tex]\(x = -20\)[/tex] is unreasonable.
