Question

In the derivation of the quadratic formula by completing the square, the equation (x+b over 2a)^2=-4ac+b^2 over 4a^2 is created by forming a perfect square trinomial. What is the result of applying the square root property of equality to this equation?

Answer 1

Answer:

The result of applying the square root property of equality to this equation is $x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$.

Step-by-step explanation:

Consider the provided equation.

$\left(x+\dfrac{b}{2a}\right)^2=\dfrac{-4ac+b^2}{4a^2}$

As the above equation is formed by perfect square trinomial so simply applying the square root property as shown:

$\sqrt{(x+\dfrac{b}{2a})^2}=\pm \dfrac{\sqrt{-4ac+b^2}}{\sqrt{4a^2}}\\x+\dfrac{b}{2a}=\pm \dfrac{\sqrt{b^2-4ac}}{2a}$

Isolate the variable x.

$x=-\dfrac{b}{2a}\pm \dfrac{\sqrt{b^2-4ac}}{2a}\\x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

Hence, the result of applying the square root property of equality to this equation is $x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$.

Answer 2

Answer:

$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

Step-by-step explanation:

The given equation is

$(x+\dfrac{b}{2a})^2=\dfrac{-4ac+b^2}{4a^2}$

In the derivation of the quadratic formula by completing the square, the above equation is created by forming a perfect square trinomial.

Applying the square root property of equality to this equation, we get

$\sqrt{(x+\dfrac{b}{2a})^2}=\pm \sqrt{\dfrac{-4ac+b^2}{4a^2}}$

$x+\dfrac{b}{2a}=\pm \dfrac{\sqrt{-4ac+b^2}}{\sqrt{4a^2}}$

$x+\dfrac{b}{2a}=\pm \dfrac{\sqrt{b^2-4ac}}{2a}$

Subtract $\frac{b}{2a}$ from both sides.

$x+\dfrac{b}{2a}-\dfrac{b}{2a}=\pm \dfrac{\sqrt{b^2-4ac}}{2a}-\dfrac{b}{2a}$

$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

Therefore, the required equation is $x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$.