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Find the value of x?
How to solve √x-1 + √x+2 = 3

In this video, we'll show you How to Solve Math Olympiad Question A Nice Radical Equation √x-1 + √x+2 = 3 in a clear , fast and easy way. Whether you are a student learning basics or a professtional looking to improve your skills, this video is for you. By the end of this video, you'll have a solid understanding of how to solve math olympiad exponential equations and be able to apply these skills to a variety of problems.
#matholympiad #maths #math #algebra

[digbycrankshaft]

Sqrt(x-1)=3-sqrt(x+2), square both sides, separate terms with x on one side, square both sides, x=2. Much quicker.

[CharlesChen-elot]

3:59 Mutiply folloeing at two side
(x+2)^1/2 -(x-1)^1/2

3=3* ((x+2)^1/2-(x-1)^1/2)

((x+2)^1/2-(x-1)^1/2'= 1
2(x+2)^1/2 = 4
x + 2 = 4
x = 2

[허공]

=> √(x - 1) + √(x + 2) = 3 ---(1), multiplying √(x - 1) - √(x + 2) to both sides
=> [√(x - 1) + √(x + 2)] * [√(x - 1) - √(x + 2)] = 3 * [√(x - 1) - √(x + 2)]
=> (x - 1) - (x + 2) = -3 = 3[√(x - 1) - √(x + 2)] => √(x - 1) - √(x + 2) = -1 ---(2)
=> (1) + (2) => 2√(x - 1) = 2 => √(x - 1) = 1 => x = 2

[ChavoMysterio]

√(x-1)+√(x+2)=3
3-√(x-1)=√(x+2)

Domain: 1≤x≤10

3-√(x-1)=√(x-1+3)

Let y=x-1

3-√y=√(y+3)

Let z=√y, z²=y

3-z=√(z²+3)
z²-6z+9=z²+3
9-6z=3
-6z=-6
x=1
√y=1
y=1
x-1=1
x=2 ❤

[seedseed]

let A= root x-1, B= root AA-BB=(A+B)(A-B)=-3 -> A-B=-1, A=1, B=2

[davidshen]

A=Sqrt（X+2）， B=Sqrt（X-1），A+B=3， A^2-B^2=3， A-B=1，A=2，B=1，X=2。

[prollysine]

let u=x-1 --> x+2=u+3, (V(u+3))^2=(3-Vu)^2, u+3=9-6Vu+u, 6Vu=6, Vu=1, (Vu)^2=1^2, u=1, x-1=1, solu, x=2,

[walterwen]

[√(x + 2)]² = [3 – √(x – 1)]², x + 2 = 9 – 6√(x – 1) + (x – 1), 6√(x – 1) = 6
√(x – 1) = 1, x – 1 = 1² = 1; x = 2

[LITHICKROSHANMS-gwlx]

My solution

√(x-1)+√(x+2)=3
√(x-1)+√(x+2)=1+2
√(x-1)+√(x+2)=√(1)+√(4)
Modifying the structure of RHS
√(x-1)+√(x+2)=√(2-1)+√(2+2)
Now comparing the structure of LHS & RHS
We get,
x=2

[dqjfrue]

Math Olympiad?
What's the "olympiadness" of the problem? It's solvable even without a pen.

[ConradoPeter-hlij]

(x-1)½+(x+2)½=3; make it:
u=(x-1)½ and v=(x+2)½
then,
u+v=3
(u+v)(u-v)=3(u-v)
u²-v²=3(u-v);
but u²=(x-1) and v²=x+2
then,
(x-1)-(x+2)= 3(u-v)
-3=3(u-v)
u-v=-1

make the sistem:
u+v=3
u-v=-1

then,
2u=2
therefore,
u=1 and v=2

But u=(x-1)½ and v=(x+2)½
then,
(x-1)½=1 and (x+2)½=2
therefore,
x-1=1 => x=2
and
x+2=4 =>x=2

therefore,
x=2 is a solution.

/ /
simplify one term way:
(x-1)½+(x+2)½=3; u=x+2 =>x-1=u-3
(u-3)½+u½=3
(u-3)½=3-u½
(u-3)½=(-1)[(u½)-3]
[(u-3)½]²={(-1)[(u½)-3]}²
u-3=(-1)²[(u½)-3]²
u-3=[(u½)-3]²
u-3=(u½)²-2*(u½)*3+3²
u-3=u-6(u½)+9
6(u½)=12
u½=2
u=4
but u=x+2; then 4=x+2.
therefore x=2

/ /
cheating way: (knowing √1=1)
(x-1)½+(x+2)½=3
when x-1=1, then (x-1)½=1 because
√1=1. So,
1+√(2+2)=3
1+√4=3
1+2=3 true.
therefore x=2 is a solution.

/ /
algebrical way:
(x-1)½+(x+2)½=3; make it:
a=(x-1)½ and b=(x+2)½
then
a+b=3
b = 3-a
b(3+a) = (3-a)(3+a)
3b+ab=3²-a²; but 3=a+b
(a+b)b+ab=9-a²
ab+b²+ab=9-a²
2ab=9-(a²+b²)
2√(x-1)√(x+2)=9-[(x-1)+(x+2)]
2√[(x-1)(x+2)]=9-[2x+1]
2√[x²+(-1+2)x+(-1)(2)]=9-2x-1
2√[x²+x-2]=8-2x
2(x²+x-2)½=2(4-x)
(x²+x-2)½=(4-x); make it:
c=(x²+x-2)½ and b=(4-x)
c=d
c*d=d*d
cd=d², but c=d; then cd=c²
therefore,
c²=d²
[(x²+x-2)½]²=(4-x)²
x²+x-2=(4²-2*4*x+x²)
x²+x-2=(16-8x+x²)
x²+(x-2)=x²+(16-8x)
(x-2)=(16-8x)
(x-2)=8(2-x)
(x-2)=8(-1)(x-2)
(1)(x-2)=(-8)(x-2); z=x-2
[1)*z=(-8)*z is true only if: z=0
then, x-2=0
x=2 is a solution.