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Sagot :
Certainly! Let's solve the quadratic equation [tex]\( x^2 + 3x - 54 = 0 \)[/tex] step-by-step.
### Step 1: Write down the quadratic equation
[tex]\[ x^2 + 3x - 54 = 0 \][/tex]
### Step 2: Identify the coefficients
The standard form of a quadratic equation is [tex]\( ax^2 + bx + c = 0 \)[/tex]. From the given equation, we identify the coefficients:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = 3 \)[/tex]
- [tex]\( c = -54 \)[/tex]
### 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]\[ \Delta = 3^2 - 4(1)(-54) \][/tex]
[tex]\[ \Delta = 9 + 216 \][/tex]
[tex]\[ \Delta = 225 \][/tex]
### Step 4: Determine the roots using the quadratic formula
The quadratic formula is:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
Substitute [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(\Delta\)[/tex] into the formula:
[tex]\[ x = \frac{-3 \pm \sqrt{225}}{2(1)} \][/tex]
[tex]\[ x = \frac{-3 \pm 15}{2} \][/tex]
### Step 5: Calculate the two possible solutions
1. For the positive square root:
[tex]\[ x = \frac{-3 + 15}{2} = \frac{12}{2} = 6 \][/tex]
2. For the negative square root:
[tex]\[ x = \frac{-3 - 15}{2} = \frac{-18}{2} = -9 \][/tex]
### Step 6: Write down the solutions
The solutions to the equation [tex]\( x^2 + 3x - 54 = 0 \)[/tex] are:
[tex]\[ x = -9 \quad \text{and} \quad x = 6 \][/tex]
So, the solutions are [tex]\( x = -9 \)[/tex] and [tex]\( x = 6 \)[/tex].
### Step 1: Write down the quadratic equation
[tex]\[ x^2 + 3x - 54 = 0 \][/tex]
### Step 2: Identify the coefficients
The standard form of a quadratic equation is [tex]\( ax^2 + bx + c = 0 \)[/tex]. From the given equation, we identify the coefficients:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = 3 \)[/tex]
- [tex]\( c = -54 \)[/tex]
### 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]\[ \Delta = 3^2 - 4(1)(-54) \][/tex]
[tex]\[ \Delta = 9 + 216 \][/tex]
[tex]\[ \Delta = 225 \][/tex]
### Step 4: Determine the roots using the quadratic formula
The quadratic formula is:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
Substitute [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(\Delta\)[/tex] into the formula:
[tex]\[ x = \frac{-3 \pm \sqrt{225}}{2(1)} \][/tex]
[tex]\[ x = \frac{-3 \pm 15}{2} \][/tex]
### Step 5: Calculate the two possible solutions
1. For the positive square root:
[tex]\[ x = \frac{-3 + 15}{2} = \frac{12}{2} = 6 \][/tex]
2. For the negative square root:
[tex]\[ x = \frac{-3 - 15}{2} = \frac{-18}{2} = -9 \][/tex]
### Step 6: Write down the solutions
The solutions to the equation [tex]\( x^2 + 3x - 54 = 0 \)[/tex] are:
[tex]\[ x = -9 \quad \text{and} \quad x = 6 \][/tex]
So, the solutions are [tex]\( x = -9 \)[/tex] and [tex]\( x = 6 \)[/tex].
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