Answered

Welcome to Westonci.ca, where finding answers to your questions is made simple by our community of experts. Connect with professionals on our platform to receive accurate answers to your questions quickly and efficiently. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.

In the derivation of the quadratic formula by completing the square, the equation [tex]\(\left(x+\frac{b}{2a}\right)^2=\frac{-4ac+b^2}{4a^2}\)[/tex] is created by forming a perfect square trinomial.

What is the result of applying the square root property of equality to this equation?

A. [tex]\(x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=\frac{-4ac+b^2}{4a^2}\)[/tex]

B. [tex]\(x+\frac{b}{2a}=\frac{\pm \sqrt{b^2-4ac}}{2a}\)[/tex]

C. [tex]\(x+\frac{b}{2a}=\frac{b \pm \sqrt{-4ac}}{2a}\)[/tex]

D. [tex]\(\left(x+\frac{b}{2a}\right)^4=\left(\frac{-4ac+b^2}{4a^2}\right)^2\)[/tex]

Sagot :

GinnyAnswer
To solve the quadratic equation by completing the square, we start with the general form of a quadratic equation:

[tex]\[ ax^2 + bx + c = 0 \][/tex]

Let's complete the square for the quadratic term. First, we divide the entire equation by [tex]\( a \)[/tex] to simplify:

[tex]\[ x^2 + \frac{b}{a}x + \frac{c}{a} = 0 \][/tex]

Next, we move the constant term ([tex]\( \frac{c}{a} \)[/tex]) to the other side of the equation:

[tex]\[ x^2 + \frac{b}{a}x = -\frac{c}{a} \][/tex]

To complete the square, we add and subtract the square of half the coefficient of [tex]\( x \)[/tex] inside the equation. The coefficient of [tex]\( x \)[/tex] is [tex]\( \frac{b}{a} \)[/tex]. Half of this coefficient is [tex]\( \frac{b}{2a} \)[/tex], and squaring it gives [tex]\( \left( \frac{b}{2a} \right)^2 = \frac{b^2}{4a^2} \)[/tex]. We add and subtract this term:

[tex]\[ x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} - \frac{b^2}{4a^2} = -\frac{c}{a} \][/tex]

Simplifying the left side, we get a perfect square trinomial:

[tex]\[ \left( x + \frac{b}{2a} \right)^2 - \frac{b^2}{4a^2} = -\frac{c}{a} \][/tex]

Now, we move the negative squared term to the right side of the equation:

[tex]\[ \left( x + \frac{b}{2a} \right)^2 = \frac{b^2}{4a^2} - \frac{c}{a} \][/tex]

Finding a common denominator for the terms on the right side:

[tex]\[ \left( x + \frac{b}{2a} \right)^2 = \frac{b^2 - 4ac}{4a^2} \][/tex]

Next, apply the square root property to both sides to isolate [tex]\( x \)[/tex]:

[tex]\[ x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a} \][/tex]

Finally, solve for [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]

Rearranging the terms under a common denominator:

[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

By applying this property to our specific problem, we get the following results:

[tex]\[ \left( x + \frac{b}{2a} \right) = \frac{\sqrt{b^2 - 4ac}}{2a} \][/tex]

and

[tex]\[ \left( x + \frac{b}{2a} \right) = -\frac{\sqrt{b^2 - 4ac}}{2a} \][/tex]

Therefore, the result of applying the square root property of equality to the given equation is:

[tex]\[ \text{Eq}\left(x + \frac{b}{2a}, \frac{\sqrt{b^2 - 4ac}}{2a}\right) \][/tex]

and

[tex]\[ \text{Eq}\left(x + \frac{b}{2a}, -\frac{\sqrt{b^2 - 4ac}}{2a}\right) \][/tex]
We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.