Certainly! Let's prove that the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] has roots given by [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex].
To do this, we will derive the quadratic formula step-by-step.
### Step 1: Write the General Form
Consider the quadratic equation:
[tex]\[ ax^2 + bx + c = 0 \][/tex]
### Step 2: Normalize the Equation
If [tex]\( a \neq 0 \)[/tex], we can divide the entire equation by [tex]\( a \)[/tex] to simplify:
[tex]\[ x^2 + \frac{b}{a} x + \frac{c}{a} = 0 \][/tex]
### Step 3: Complete the Square
To complete the square, we need to make the left side of the equation a perfect square trinomial. Let's isolate the constant term:
[tex]\[ x^2 + \frac{b}{a} x = -\frac{c}{a} \][/tex]
Next, add and subtract [tex]\((\frac{b}{2a})^2\)[/tex] on the left-hand side:
[tex]\[ x^2 + \frac{b}{a} x + \left(\frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2 = -\frac{c}{a} \][/tex]
This simplifies to:
[tex]\[ \left(x + \frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2 = -\frac{c}{a} \][/tex]
### Step 4: Simplify the Equation
Combine the terms involving [tex]\( x \)[/tex]:
[tex]\[ \left(x + \frac{b}{2a}\right)^2 = \frac{b^2}{4a^2} - \frac{c}{a} \][/tex]
Now, get a common denominator on the right-hand side and simplify:
[tex]\[ \left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2} \][/tex]
### Step 5: Solve for [tex]\( x \)[/tex]
Take the square root of both sides:
[tex]\[ x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a} \][/tex]
Isolate [tex]\( x \)[/tex] by subtracting [tex]\(\frac{b}{2a}\)[/tex] from both sides:
[tex]\[ x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a} \][/tex]
### Step 6: Combine the Terms
Combine the terms on the right-hand side under a common denominator:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Thus, we have derived that the roots of the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] are given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
### Conclusion:
If [tex]\( a = 1 \)[/tex], [tex]\( b = -3 \)[/tex], and [tex]\( c = 2 \)[/tex], then substituting these values into the quadratic formula gives:
[tex]\[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(2)}}{2(1)} \][/tex]
[tex]\[ x = \frac{3 \pm \sqrt{9 - 8}}{2} \][/tex]
[tex]\[ x = \frac{3 \pm 1}{2} \][/tex]
Thus, the roots are:
[tex]\[ x_1 = \frac{3 + 1}{2} = 2 \][/tex]
[tex]\[ x_2 = \frac{3 - 1}{2} = 1 \][/tex]
The discriminant [tex]\( \Delta \)[/tex] here is [tex]\( b^2 - 4ac = 1 \)[/tex], confirming that the roots indeed are [tex]\( 2.0 \)[/tex] and [tex]\( 1.0 \)[/tex].
