Solve <span title="Sqroot(4y-9)-sqroot(5y-4)=1?">Sqroot(4y-9)-sqroot(5y-4)...</span>

Question

I have no clue on solving this...

Solve Sqroot(4y-9)-sqroot(5y-4)=1

Answer 1

sqrt(4y - 9) = 1 + sqrt(5y - 4)

4y - 9 = 1 + 2 sqrt(5y - 4) + 5y - 4

-(y + 6) = 2 sqrt(5y - 4)

y^2 + 12y + 36 = 4(5y -4)

y^2 - 8y + 52 = 0

Now use the quadratic formula to get

y = 4 + i sqrt 3, 4 - i sqrt 3

you have a pair of complex conjugate solutions.

Answer 2

There is no real solution for this problem.

Since √(4y-9) - √(5y-4) = 1 then

√(4y-9) &gt; √(5y-4)

4y - 9 &gt; 5y - 4

4y - 5y &gt; 9 - 4

-y &gt; 5

y &lt; -5

But when y &lt; -5 the values inside the square roots are negative.

Answer 3

To solve this, you want to get one root on one side, then square:

√(4Y-9) - √(5Y-4) = 1

Add the second root to both sides:

√(4Y-9) = 1 + √(5Y-4)

Square both sides:

4Y-9 = 1 + 2√(5Y-4) + 5Y-4

Repeat the process here to get √(5Y-4) as the only thing on the right.

Square both sides again.

Now, you can solve quadratically.