How do you simplify #sqrt(x-1) + sqrt( 2x) = 3#?

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Answer 1 · 2018-05-06T16:13:34

#rarrx=2#

#rarrsqrt(x-1)+sqrt(2x)=3#

#rarrsqrt(x-1)=3-sqrt(2x)#

#rarr[sqrt(x-1)]^2=[3-sqrt(2x)]^2#

#rarrx-1=9-6sqrt(2x)+2x#

#rarr6sqrt(2x)=x+10#

#rarr[6sqrt(2x)]^2=[x+10]^2#

#rarr36*(2x)=x^2+20x+100#

#rarrx^2-52x+100=0#

#rarrx^2-2*x*26+26^2-26^2+100=0#

#rarr(x-26)^2=26^2-100=576#

#rarrx-26=sqrt(576)=+-24#

#rarrx=26+24,26-24=50 or 2#

Putting #x=50# in given equation, we get,

#rarrsqrt(50-1)+sqrt(2*50)=17(rejected)#

Putting #x=2# in given equation, we get,

#rarrsqrt(2-1)+sqrt(2*2)=3(accepted)#

So, the required value of x is #2.#