You have solved equations that involve squareroot s of algebraic expressions. Th

Question

You have solved equations that involve squareroot s of algebraic expressions. Then you checked your solutions to see if they are actual solutions or extraneous solutions. To find the answer to the question at the bottom of the page: Solve each equation Use the space below to show your work Write solutions below each equation Identify any extraneous solutions. Use only the actual solutions that are positive integers to answer the question x = Squareroot 4x - 2 x = Squareroot 3 - 2x x = Squareroot -2x + 48 x = Squareroot 90 - x

Explanation / Answer

a) x = sqrt(4x -2)

x^2 = 4x -2

x^2 - 4x + 2=0

x= (4 +/- sqrt(16 -8) )/2 = 1/2 +/- sqrt2

in original equation : check roots for sqrt(4x -2)

x = sqrt( 4(1/2 +/- sqrt2) -2 ) = sqrt( 2 +/- 4sqrt2 -2) = sqrt(+/-4sqrt2)

x = 1/2 - sqrt2 is extraneous solution

b) x = sqrt( 3 -2x)

x^2 = 3 -2x

x^2 +2x -3 =0

x = [ -2 +/- sqrt( 4 +12) ]/2 = -1 +/-2 = -3 , 1

x = 1, -3 Plug these solution in original equation

Neglect x = -3 as mentioined in the question and this is the extraneous solution

Only solution x= 1

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