Question 164396
{{{sqrt(x)+ 4 = 2*sqrt(x+5)}}}
{{{(sqrt(x)+ 4)^2 = (2*sqrt(x+5))^2}}}
{{{x+8*sqrt(x)+ 16 = 4*(x+5)}}}
{{{x+8*sqrt(x)+ 16 = 4x+20}}}
{{{-3x+8*sqrt(x)-4=0}}}
{{{3x-8*sqrt(x)+4=0}}}
Look similar to a quadratic equation, let's use a substitution.
Let {{{x=u^2}}}.
{{{3u^2-8u+4=0}}}
This quadratic equation can be factored.
{{{(3u-2)(u-2)=0}}}
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First solution
{{{3u-2=0}}}
{{{3u=2}}}
{{{u=2/3}}}
{{{highlight(x=u^2=4/9)}}}
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Second solution
{{{u-2=0}}}
{{{u=2}}}
{{{highlight(x=u^2=4)}}}
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Let's check the answers.
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First solution: x=4/9
{{{sqrt(x)+ 4 = 2*sqrt(x+5)}}}
{{{sqrt(4/9)+4=2*sqrt(4/9+5)}}}
{{{2/3+4=2*sqrt(4/9+45/9)}}}
{{{2/3+12/3=2*sqrt(49/9)}}}
{{{14/3=2*7/3}}}
{{{14/3=14/3}}}
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Second solution: x=4
{{{sqrt(x)+ 4 = 2*sqrt(x+5)}}}
{{{sqrt(4)+ 4 = 2*sqrt(4+5)}}}
{{{2+4=2*sqrt(9)}}}
{{{6=2*3}}}
{{{6=6}}}
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Both answers have been checked.