Certainly! Let's solve the given equation step by step.
### Step 1: Write the equation in standard quadratic form
Given equation:
[tex]\[ x(20 + x) = 125 \][/tex]
First, expand and rewrite the equation:
[tex]\[ x \cdot 20 + x \cdot x = 125 \][/tex]
[tex]\[ 20x + x^2 = 125 \][/tex]
To obtain the standard quadratic form [tex]\( ax^2 + bx + c = 0 \)[/tex], we have to move all terms to one side of the equation:
[tex]\[ x^2 + 20x - 125 = 0 \][/tex]
### Step 2: Identify coefficients
From the quadratic equation [tex]\( x^2 + 20x - 125 = 0 \)[/tex], we identify the coefficients:
- [tex]\( a = 1 \)[/tex] (coefficient of [tex]\( x^2 \)[/tex] )
- [tex]\( b = 20 \)[/tex] (coefficient of [tex]\( x \)[/tex] )
- [tex]\( c = -125 \)[/tex] (constant term)
### Step 3: Calculate the discriminant
The discriminant [tex]\( \Delta \)[/tex] of a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substitute the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ \Delta = 20^2 - 4 \cdot 1 \cdot (-125) \][/tex]
[tex]\[ \Delta = 400 + 500 \][/tex]
[tex]\[ \Delta = 900 \][/tex]
### Step 4: Use the quadratic formula
The quadratic formula to find the solutions of the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
Substitute [tex]\( a = 1 \)[/tex], [tex]\( b = 20 \)[/tex], and [tex]\( \Delta = 900 \)[/tex]:
[tex]\[ x = \frac{-20 \pm \sqrt{900}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-20 \pm 30}{2} \][/tex]
### Step 5: Calculate the two possible solutions
First possible solution [tex]\( x_1 \)[/tex]:
[tex]\[ x_1 = \frac{-20 + 30}{2} \][/tex]
[tex]\[ x_1 = \frac{10}{2} \][/tex]
[tex]\[ x_1 = 5 \][/tex]
Second possible solution [tex]\( x_2 \)[/tex]:
[tex]\[ x_2 = \frac{-20 - 30}{2} \][/tex]
[tex]\[ x_2 = \frac{-50}{2} \][/tex]
[tex]\[ x_2 = -25 \][/tex]
### Step 6: Present the solutions
The discriminant calculated is 900, and the solutions to the equation [tex]\( x(20 + x) = 125 \)[/tex] are:
[tex]\[ x_1 = 5 \][/tex]
[tex]\[ x_2 = -25 \][/tex]
So, the equation [tex]\( x(20 + x) = 125 \)[/tex] has two solutions: [tex]\( x = 5 \)[/tex] and [tex]\( x = -25 \)[/tex].
