Answered

At Westonci.ca, we make it easy to get the answers you need from a community of informed and experienced contributors. Join our Q&A platform and connect with professionals ready to provide precise answers to your questions in various areas. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.

Solving Radical Equations

Square root x+7 = x - 5

Sagot :

apologiabiology
so the square root of the quatity(x+7)=x-5

so you square both sides and get rid of the square root
x+7=(x-5)^2
(x-5)^2=x^2-10x+25
x+7=x^2-10x+25
subtract 7 from both sides
x=x^2-10x+18
subtract x from both sides
0=x^2-11x+18

so if xy=0 we can assume that x or/and y =0

factor out x^2-11+18
(find what two numbers multiply to get 18 and add to get -11)
-2 times -9=18
-2+(-9)=-11

(x-2)(x-9)=0
set them to zero
x-2=0
x=2

x-9=0
x=9

there are two answers -2 and -9
naǫ
The domain:
The radicand must be greater than or equal to 0.
[tex]x+7 \geq 0 \\ x \geq -7[/tex]
The value of the square root must be greater than or equal to 0.
[tex]x-5 \geq 0 \\ x \geq 0[/tex]
Therefore x≥5.

[tex]\sqrt{x+7}=x-5 \\ (\sqrt{x+7})^2=(x-5)^2 \\ x+7=x^2-10x+25 \\ 0=x^2-11x+18 \\ 0=x^2-9x-2x+18 \\ 0=x(x-9)-2(x-9) \\ 0=(x-2)(x-9) \\ x-2=0 \ \lor \ x-9=0 \\ x=2 \ \lor \ x=9[/tex]
2<5 so it's not a correct solution.
9≥5 so it's a correct solution.

The answer:
x=9
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.