Question 183287
Are you certain that the equation is not {{{sqrt(4-7x) = sqrt(2)*x}}} (the last "x" is NOT in the square root)


If so, then... 



{{{sqrt(4-7x) = sqrt(2)*x}}} Start with the given equation.



{{{4-7x = (sqrt(2)*x)^2}}} Square both sides



{{{4-7x = 2x^2}}} Square {{{sqrt(2)*x}}} to get {{{2x^2}}}



{{{-2x^2-7x+4=0}}} Subtract {{{2x^2}}} from both sides and rearrange the terms.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=-2}}}, {{{b=-7}}}, and {{{c=4}}}



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(-7) +- sqrt( (-7)^2-4(-2)(4) ))/(2(-2))}}} Plug in  {{{a=-2}}}, {{{b=-7}}}, and {{{c=4}}}



{{{x = (7 +- sqrt( (-7)^2-4(-2)(4) ))/(2(-2))}}} Negate {{{-7}}} to get {{{7}}}. 



{{{x = (7 +- sqrt( 49-4(-2)(4) ))/(2(-2))}}} Square {{{-7}}} to get {{{49}}}. 



{{{x = (7 +- sqrt( 49--32 ))/(2(-2))}}} Multiply {{{4(-2)(4)}}} to get {{{-32}}}



{{{x = (7 +- sqrt( 49+32 ))/(2(-2))}}} Rewrite {{{sqrt(49--32)}}} as {{{sqrt(49+32)}}}



{{{x = (7 +- sqrt( 81 ))/(2(-2))}}} Add {{{49}}} to {{{32}}} to get {{{81}}}



{{{x = (7 +- sqrt( 81 ))/(-4)}}} Multiply {{{2}}} and {{{-2}}} to get {{{-4}}}. 



{{{x = (7 +- 9)/(-4)}}} Take the square root of {{{81}}} to get {{{9}}}. 



{{{x = (7 + 9)/(-4)}}} or {{{x = (7 - 9)/(-4)}}} Break up the expression. 



{{{x = (16)/(-4)}}} or {{{x =  (-2)/(-4)}}} Combine like terms. 



{{{x = -4}}} or {{{x = 1/2}}} Simplify. 



So the <i>possible</i> answers are {{{x = -4}}} or {{{x = 1/2}}} 


  
However, if we plug in {{{x = -4}}}, the left side will have a negative number inside the square root. 



Since you CANNOT take the square root of a negative number, this means that we must ignore this value.



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Answer:


So the only solution is {{{x = 1/2}}}