How do you solve #2x=sqrt[12x-5]# and find any extraneous solutions?

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Answer 1 · 2017-11-09T22:11:44

Start by squaring both sides...

...giving #4x^2 = 12x - 5#

...now you can subtract 12x and add 5 to both sides, giving:

#4x^2 - 12x + 5 = 0#

...this is a quadratic equation with coefficients 4, -12, and 5.

Remember your forumula, if #ax^2 + bx + c = 0#, then your roots are:

#x = (-b +- sqrt(b^2 - 4ac))/(2a)#

...and for coefficients a = 4, b = -12, c = 5, your roots are:

2.5 and 0.5

GOOD LUCK