Question
Solve $\sqrt{y+1}+\sqrt{2 y-5}=3$
✓
Answer
$\begin{aligned} & \sqrt{y+1}+\sqrt{2 y-5}=3 \ldots \text { (squaring on both sides) } \\ & (\sqrt{y+1}+\sqrt{2 y-5})^2=3^2 \\ & (\sqrt{y+1})^2+(\sqrt{2 y-5})^2+2(\sqrt{y+1})(\sqrt{2 y-5})=9 \\ & y+1+2 y-5+2 \sqrt{(y+1)(2 y-5)}=9 \\ & 3 y-4+2 \sqrt{2 y^2-5 y+2 y-5}=9 \\ & 2 \sqrt{2 y^2-3 y-5}=9+4-3 y \\ & =13-3 y \quad \ldots \text { (squaring on both sides) } \\ & {\left[2 \sqrt{2 y^2-3 y-5}\right]^2=(13-3 y)^2} \\ & 4\left(2 y^2-3 y-5\right)=169+9 y^2-78 y \\ & 8 y^2-12 y-20=169+9 y^2-78 y \\ & 8 y^2-9 y^2-12 y+78 y-20-169=0 \\ & -y^2-66 y-189=0 \\ & \end{aligned}$
$y^2 – 66y + 189 = 0$
(y – 3) (y – 63) = 0
y – 3 or y = 63
The value of y is 3 and 63
$y^2 – 66y + 189 = 0$
(y – 3) (y – 63) = 0
y – 3 or y = 63
The value of y is 3 and 63
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