Question

Solve the equation: (√2r+11)-(√5r+1)=-1

Answer 1

\[-\sqrt{2r}-\sqrt{5r}=-11\]
multiply both sides with -1
and you will get
\[\sqrt{2r}+\sqrt{5r}=11\]

Answer 2

taking square on both sides
\[2r + 2\sqrt{2r}\sqrt{5r}+5r =121\]
\[7r+2\sqrt{2r}\sqrt{5r} = 121\]

Answer 3

7r+2r sqrt10 =121
this is the solution of it

Answer 4

or
sqrt 10= 121-7r \2r

Answer 5

The answer in the back of the textbook says it is 7. The first step that the example shows is adding (√5r+1) to both sides, then I squared both of them and got: 2r+11=5r+3(-2√5r-1). Does this way look familiar to you? It's new to me so I'm having a hard time with it.

Answer 6

then equation becomes
\[\sqrt{2r +11}=-1+\sqrt{5r+1}\]
taking sqroot onn both sides
it becomes
2r+11= -1 -2sqrt5r+1 + 25r^2 +10r +1
\[-2\sqrt{5r+1}+25r ^{2}+8r -11\]=0
now you can solve it
25r^2 +8r -11=2sqrt5r+1
first take Factorized on left hand side

Answer 7

you have\[\sqrt{2r +11}-\sqrt{5r+1}=-1\]and id probably say always for questions like this you should eliminate radicals, first isolate one of radicals, for example\[\sqrt{2r +11}=-1+\sqrt{5r+1}\]square both sides u will get\[2r+11=1+5r+1-2\sqrt{5r+1}\]simplify\[-3r+9=-2\sqrt{5r+1}\]square both sides again\[9r^2+81-54r=4(5r+1)\]simplify\[9r^2-74r+77=0\]its a qudratic with roots\[r=7\]\[r=\frac{11}{9}\]you have not considered restrictions yet so check the answers you got

there is only one solution \(r=7\)