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square root 5x^2 -3x =2x solve each radical equation and check the answer help
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\[\sqrt{5x^2-3x}=2x\] ?
yes
square both sides \[5x^2-3x=(2x)^2\] \[5x^2-3x=2^2*x^2\] \[5x^2-3x=4x^2\] subtract 4x^2 on both sides \[5x^2-4x^2-3x=4x^2-4x^2\] \[x^2-3x=0\] now factor \[x(x-3)=0\]
\[\sqrt{5x ^{2}-3x}=2x\] \[5x ^{2} -3x =4x ^{a}\] 5x^2-4x^2-3x=0 x(x-3)=0 x=0,x=3
x=0 or x=3 but since we squared both sides lets check our answers we see for 0 it is true it is true for x=3 \[\sqrt{5*3^2-3*3}=\sqrt{45-9}=\sqrt{36}=6\] other side we have 3*2=6 so both sides are the same :)x=0 and x=3 are both solutions
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