1. ## help pls

Solve the following equations .

(1)$\displaystyle \sqrt{x-2}+\sqrt{x+3}=5$

(2)$\displaystyle \sqrt{2x+11}+\sqrt{x+2}=\sqrt{5x+17}$

2. 1.sqrt(x-2)+sqrt(x+3)=5
squaring both thr sides we have(sqrt(x-2)+sqrt(x+3))^2=25
using (a+b)^2 formula i.e a^2+ b^2+2ab here we get
x-2 +x+3 +2(sqrt(x-2)sqrt(x+3)0=25
2x+1-25 = -2sqrt(x-2)sqrt(x+3)
2(x-12)=-2sqrt(x-2)sqrt(x+3)
x-6 = -sqrt(x-2)sqrt(x+3)
taking square both the sides we have
(x-6)^2 = (-sqrt(x-2)sqrt(x+3))^2
x^2-12x+36 = (x-2)(x+3)
x^2-12x+36= x^2+x-6
taking the variables to the l.h.s and constants to r.h.s we have
-12x-x=-36-6
-13x = -42
-13x/-13= -42/-13
so,x =4
the next sum you can try similarly

3. Originally Posted by sunitaparija
1.sqrt(x-2)+sqrt(x+3)=5
squaring both thr sides we have(sqrt(x-2)+sqrt(x+3))^2=25
using (a+b)^2 formula i.e a^2+ b^2+2ab here we get
x-2 +x+3 +2(sqrt(x-2)sqrt(x+3)0=25
2x+1-25 = -2sqrt(x-2)sqrt(x+3)
2(x-12)=-2sqrt(x-2)sqrt(x+3)
x-6 = -sqrt(x-2)sqrt(x+3)
taking square both the sides we have
(x-6)^2 = (-sqrt(x-2)sqrt(x+3))^2
x^2-12x+36 = (x-2)(x+3)
x^2-12x+36= x^2+x-6
taking the variables to the l.h.s and constants to r.h.s we have
-12x-x=-36-6
-13x = -42
-13x/-13= -42/-13
so,x =4
the next sum you can try similarly

I guess you made a mistake somewhere
$\displaystyle \sqrt{x-2}+\sqrt{x+3}=5$
$\displaystyle 2x+1+2\sqrt{x-2}\sqrt{x+3}=25$
$\displaystyle (-x+12)^2=(2\sqrt{x-2}\sqrt{x+3})^2$
$\displaystyle 144-24x+x^2=(x-2)(x+3)$
and i finally got x=6
however , thanls for helping me out .