Mathematics: PLEASE help me with this question!

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PLEASE help me with this question! - id7167660, exiZT7535

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playboib

30.01.2022

Mathematics

$\huge\boxed{x = 8,\ but\ x = 3\ is\ extraneous}$

Step-by-step explanation:

$\sqrt{x+1}+5=x\\\\\text{Domain:}\ x+1\geq0\to x\geq-1\\\\\sqrt{x+1}+5=x\qquad\text{subtract 5 from both sides}\\\sqrt{x+1}=x-5\qquad\text{square both sides}\\(\sqrt{x+1})^2=(x-5)^2\qquad\text{use}\ (\sqrt{x})^2=x\ \text{and}\ (a-b)^2=a^2-2ab+b^2\\x+1=x^2-2(x)(5)+5^2\\x+1=x^2-10x+25\qquad\text{subtract}\ x\ \text{and}\ 1\ \text{from both sides}\\0=x^2-11x+24$

$ab=24;\ a+b=-11\\a=-8;\ b=-3\\\\x^2-11x+24=0\\x^2-8x-3x+24=0\qquad\text{distribute}\\x(x-8)-3(x-8)=0\qquad\text{distribute}\\(x-8)(x-3)=0\\\\\text{The product is 0 if one of the factors is 0. Therefore}\\(x-8)(x-3)=0\iff x-8=0\ \vee\ x-3=0\\x-8=0\qquad\text{add 8 to both sides}\\\boxed{x=8}\in D\\x-3=0\qquad\text{add 3 to both sides}\\\boxed{x=3}\in D$

$\text{For}\ x=8:\\\sqrt{8+1}+5=8\\\sqrt{9}+5=8\\3+5=8\\8=8\qquad\bold{CORRECT}\\\text{For}\ x=3:\\\sqrt{3+1}+5=3\\\sqrt4+5=3\\2+5=3\\7=3\qquad\bold{FALSE}$

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