How do you solve #\sqrt { 2a + 5} - 2\sqrt { 2a } = 1#?

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Answer 1 · 2018-08-05T09:46:57

#a=2/9#

Here ,

#sqrt(2a+5)-2sqrt(2a)=1...to(1)#

#:.sqrt(2a+5)=1+2sqrt(2a)#

Squaring both sides of eqn.

#(sqrt(2a+5))^2=(1+2sqrt(2a))^2#

#:.2a+5=1+4sqrt(2a)+8a#

#:.2a-8a+5-1=4sqrt(2a)#

#:.4-6a=4sqrt(2a)#

#:.2-3a=2sqrt(2a)#

Again squaring both sides of eqn.

#(2-3a)^2=(2sqrt(2a))^2#

#:.4-12a+9a^2=8a#

#:.9a^2-20a+4=0#

Now ,

#(-18a)+(-2a)=-20a and (-18a)xx(-2a)=36a^2#

#9a^2-18a-2a+4=0#

#:.9a(a-2)-2(a-2)=0#

#:.(a-2)(9a-2)=0#

#:.a-2=0 or 9a-2=0#

#:.a=2 or a=2/9#

Check :

#(i)a=2#

#LHS=sqrt(2a+5)-2sqrt(2a)#

#LHS=sqrt(2(2)+5)-2sqrt(2(2))#

#:.LHS=sqrt9-2sqrt4=3-4=-1!=RHS#

#:.a!=2#

#(ii)a=2/9#

#LHS=sqrt(2a+5)-2sqrt(2a)#

#LHS=sqrt(2(2/9)+5)-2sqrt(2(2/9))#

#:.LHS=sqrt(4/9+5)-2sqrt(4/9)=7/3-4/3=3/3=1=RHS#

#:.a=2/9#