Welcome to Westonci.ca, your one-stop destination for finding answers to all your questions. Join our expert community now! Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
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].
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].
We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.