Question 836280
This is a long and messy problem. The basic idea is this though: Isolate the square roots one by one on their own side. That will allow you to square both sides to eliminate each square root one by one. After doing all that, you can use the quadratic formula to solve for x. Remember to check ALL possible answers.


{{{7-sqrt(1-4x)-sqrt(4-6x)=0}}}


{{{7-sqrt(1-4x)=sqrt(4-6x)}}}


{{{(7-sqrt(1-4x))^2=(sqrt(4-6x))^2}}}


{{{(7-sqrt(1-4x))(7-sqrt(1-4x))=4-6x}}}


{{{7(7-sqrt(1-4x))-sqrt(1-4x)(7-sqrt(1-4x))=4-6x}}}


{{{7(7)+7(-sqrt(1-4x))-sqrt(1-4x)(7)-sqrt(1-4x)(-sqrt(1-4x))=4-6x}}}


{{{49-7sqrt(1-4x)-7sqrt(1-4x)+(sqrt(1-4x))^2=4-6x}}}


{{{49-7sqrt(1-4x)-7sqrt(1-4x)+1-4x=4-6x}}}


{{{50-4x-14sqrt(1-4x)=4-6x}}}


{{{-14sqrt(1-4x)=4-6x-50+4x}}}


{{{-14sqrt(1-4x)=-46-2x}}}


{{{(-14sqrt(1-4x))^2=(-46-2x)^2}}}


{{{196(1-4x)=(-46-2x)(-46-2x)}}}


{{{196(1-4x)=-46(-46-2x)-2x(-46-2x)}}}


{{{196(1)+196(-4x)=-46(-46)-46(-2x)-2x(-46)-2x(-2x)}}}


{{{196-784x=2116+92x+92x+4x^2}}}


{{{0=2116+92x+92x+4x^2-196+784x}}}


{{{0=4x^2+968x+1920}}}


{{{4x^2+968x+1920=0}}}


{{{4(x^2+242x+480)=0}}}


{{{x^2+242x+480=0/4}}}


{{{x^2+242x+480=0}}}


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Now use the quadratic formula to solve for x


*[invoke quadratic_formula 1, 242, 480, "x"]



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So the *possible* solutions are {{{x = -2}}} or {{{x = -240}}}


However, we must check them both


Checking {{{x = -2}}}


{{{7-sqrt(1-4x)-sqrt(4-6x)=0}}}


{{{7-sqrt(1-4(-2))-sqrt(4-6(-2))=0}}}


{{{7-sqrt(1+8)-sqrt(4+12)=0}}}


{{{7-sqrt(9)-sqrt(16)=0}}}


{{{7-3-4=0}}}


{{{0=0}}} This is TRUE, so {{{x = -2}}} is definitely a solution.


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Checking {{{x = -240}}}


{{{7-sqrt(1-4x)-sqrt(4-6x)=0}}}


{{{7-sqrt(1-4(-240))-sqrt(4-6(-240))=0}}}


{{{7-sqrt(1+960)-sqrt(4+1440)=0}}}


{{{7-sqrt(961)-sqrt(1444)=0}}}


{{{7-31-38=0}}}


{{{-62=0}}} This is FALSE, so {{{x = -240}}} is NOT a solution.



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Final Answer: {{{x = -2}}}