Use a=√x, and a^2=x, then : a +√a^2-9=9 √a^2-9= 9-a (√a^2-9)^2= (9-a)^2 a^2-9= 81-18a +a^2 (cancel a^2 both members) 18a=81+9 18a=90 a=90/18 a=5. And a=√x: 5=√X X=25.
stupid solution for two square roots to add up to an integer they must be integer square roots. One is 9 less than the other. This is only the case for 25 and 16. Answer 25.
Use a=√x, and a^2=x, then
: a +√a^2-9=9
√a^2-9= 9-a
(√a^2-9)^2= (9-a)^2
a^2-9= 81-18a +a^2 (cancel a^2 both members)
18a=81+9
18a=90
a=90/18
a=5.
And a=√x:
5=√X
X=25.
I hope I helped , i am from Brazil
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asnwer-3/3 isit
asnwer=25 isit hmm fmm
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{x+x ➖ }(x)^2 ➖ (9)^2=x^2+{x^2 ➖ 81}=x^2+{x^0+x^0 ➖ x^0+x^0 ➖ }=x^2+{x^1+x^1}={x^2+x^2}=x^4"x^2^2 (x ➖ 2x+2) .
stupid solution
for two square roots to add up to an integer they must be integer square roots. One is 9 less than the other. This is only the case for 25 and 16. Answer 25.
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