Question

How do you solve `sqrt(2x+3) + sqrt(4-x) = 4`?

Answer 1

`x=9`

Explanation:

`sqrt(2x+3) + sqrt(4-x)=4`

square both sides:

`(2x+3) + (4-x)=4^2 color(red)(larr" gone astray on this line")`

expand brackets:

`2x+3 +4 -x=16`

simplify:

`x+7=16`

subtract 7:

`x=9`

Answer 2

Taken you to a point where you could take over

Explanation:

Write as: `(sqrt(2x+3)+sqrt(4-x))=4`

Square both sides:

`(color(white)(.)sqrt(2x+3)+sqrt(4-x)color(white)(.))(color(white)(.)sqrt(2x+3)+sqrt(4-x)color(white)(.))=16`

`(sqrt(2x+2))^2 +sqrt(4-x)sqrt(2x+3)+(sqrt(4-x))^2=16`

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To check how something works by example:
Consider: `sqrt(4)xxsqrt(9) = 2xx3=6`
Can we combine these?`->sqrt(4xx9)=sqrt(36)=6` YES!
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`(sqrt(2x+2))^2 +sqrt((4-x)(2x+3))+(sqrt(4-x))^2=16`

`2x+2+sqrt(-2x^2+5x+12) +4-x=16`

`sqrt(-2x^2+5x+12)=-x-10`

Square both sides

`-2x^2+5x+12=x^2+20x+100`

`3x^2+15x+88=0`

I will let you tale over from this point