Respuesta :
Answer:
The result of applying the square root property of equality to this equation is [tex]x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex].
Step-by-step explanation:
Consider the provided equation.
[tex]\left(x+\dfrac{b}{2a}\right)^2=\dfrac{-4ac+b^2}{4a^2}[/tex]
As the above equation is formed by perfect square trinomial so simply applying the square root property as shown:
[tex]\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}[/tex]
Isolate the variable x.
[tex]x=-\dfrac{b}{2a}\pm \dfrac{\sqrt{b^2-4ac}}{2a}\\x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
Hence, the result of applying the square root property of equality to this equation is [tex]x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex].
To solve the problem we must know about the square root property of equality.
when the square root property of equality is applied to the equation [tex](x+\dfrac{b}{2a})^2=\dfrac{-4ac+b^2}{4a^2}[/tex] it forms the equation of finding the roots of a quadratic equation, therefore, [tex]x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex].
What is the square root property of equality?
The square root property of equality states that if we have an equation such that one side of the equation has squared values while the other has a number then we have to take the square root of both the sides of the equation, Also put a '±' sign on the side with the number, therefore,
a² = c
√(a²) = ±√c
Given to us
[tex](x+\dfrac{b}{2a})^2=\dfrac{-4ac+b^2}{4a^2}[/tex]
We already discussed the square root property of equality, therefore,
[tex](x+\dfrac{b}{2a})^2=\dfrac{-4ac+b^2}{4a^2}\\\\\\\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{-4ac+b^2}}{2a}\\\\\\x = \dfrac{-b}{2a} \pm \dfrac{\sqrt{-4ac+b^2}}{2a}\\\\\\x = \dfrac{-b \pm \sqrt{-4ac+b^2}}{2a}\\\\\\x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]
Thus, when the square root property of equality is applied to the equation [tex](x+\dfrac{b}{2a})^2=\dfrac{-4ac+b^2}{4a^2}[/tex] it forms the function of finding the roots of a quadratic equation, therefore, [tex]x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex].
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