Question

Solve the following by reducing them to quadratic form:
`sqrt(y + 1) + sqrt(2y - 5) = 3, y ∈ "R".`

Answer 1

Given equation
`sqrt(y + 1) + sqrt(2y - 5)` = 3
⇒ `sqrt(y + 1) = 3 - sqrt(2y - 5)`
Squaring both sides, we get
y + 1 = 9 + 2y - 5 - 6`sqrt(2y - 5)`
⇒ y - 2y + 1 - 4 = -6`sqrt(2y - 5)`
-y - 3 = -6`sqrt(2y - 5)`
⇒ y + 3 = 6`sqrt(2y - 5)`
On Squaring again, we get
y² + 9 + 6y = 36(2y - 5)
⇒ y² + 9 + 6y = 72y - 180
⇒ y² + 6y - 72y + 9 + 180 = 0
⇒ y² - 66y + 189 = 0
∴ y² - 66y + 189 = 0
Hence, a = 1, b = -66, c = 189
Then, D 
= b² - 4ac
= (66)² - 4(1) (189)
= 4356 - 756
= 3600 > 0
Roots are real.
∴ y = `(-b ± sqrt(b^2 - 4ac))/(2a)`
y = `(-(-66) ± sqrt(3600))/(2 xx 1) `
y = `(66 ± 60)/(2)`
y = `(66 + 60)/(2), (66 - 60)/(2)`
= `(126)/(2), (6)/(2)`
= 63, 3
y = {63, 3}
But x = 63 does not satisfy the given equation
Hence, the solution is 3.