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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]
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