Experience the power of community-driven knowledge on IDNLearn.com. Join our interactive community and get comprehensive, reliable answers to all your questions.

Simplify surd need an answer asap please

Simplify Surd Need An Answer Asap Please class=

Sagot :

Answer:

Step-by-step explanation:

Simplify Surd:

  • Multiply the denominator and numerator with the conjugate of the denominator.
  • Conjugate of denominator:

           [tex]\sf (\sqrt{x+1}) + (\sqrt{x-1})[/tex]

[tex]\sf\dfrac{\sqrt{x+1} +\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}} = \dfrac{(\sqrt{x+1} +\sqrt{x-1})(\sqrt{x+1} +\sqrt{x-1})}{(\sqrt{x+1} -\sqrt{x-1})(\sqrt{x+1} +\sqrt{x-1})}\\[/tex]

           When doing so, in the numerator, we get in the form of the identity (a +b)² and in the denominator a² - b²

                            [tex]\sf =\dfrac{(\sqrt{x+1} +\sqrt{x-1})^2}{(\sqrt{x+1})^2 - (\sqrt{x-1})^2}\\\\=\dfrac{(\sqrt{x+1})^2+(\sqrt{x-1})^2+2*\sqrt{x+1}*\sqrt{x-1}}{x + 1 - (x - 1)}\\\\=\dfrac{x+1 +x -1 + 2*\sqrt{(x+1)(x-1)}}{x + 1 - x + 1}\\\\=\dfrac{x+ x + 1 - 1 +2\sqrt{x^2-y^2}}{x-x+1+1}\\\\=\dfrac{2x+2\sqrt{x^2-y^2}}{2}\\\\= \dfrac{2x}{2}+\dfrac{2\sqrt{x^2-y^2}}{2}\\\\=x + \sqrt{x^2-y^2}[/tex]