How do you solve #sqrt(a+21)-1=sqrt(a+12)#?

2 Answers

Answer:

the answer is #a=4#

Explanation:

the given equation is #sqrt(a+21)-1=sqrt(a+12)#
squaring on both sides we get
#a+21+1-2sqrt(a+21)=a+12#
simplifying we get #2sqrt(a+21)=10rArrsqrt(a+21)=5#
squaring on both sides we get
#a+21=25rArra=4#

Nov 8, 2017

Answer:

#a=4#

Explanation:

#sqrt(a+21)-1=sqrt(a+12)#

#sqrt(a+21)-sqrt(a+12)=1#

After using difference of squares identity,

#[(a+21)-(a+12)]/[sqrt(a+21)-sqrt(a+12)]=9/1#

#sqrt(a+21)+sqrt(a+12)=9#

Hence,

#sqrt(a+21)+sqrt(a+12)+sqrt(a+21)-sqrt(a+12)=9+1#

#2sqrt(a+21)=10#

#sqrt(a+21)=5#

#a+21=25#

#a=4#