find the roots of the equation x+ √x-2 = 4?
回答 (6)
2016-02-11 3:18 pm
If you mean: x + √(x - 2) = 4
then, isolate the radical, square both sides, solve the resulting quadratic equation, check...
√(x - 2) = (4 - x)
(x - 2) = 16 - 8x + x²
x² - 9x + 18 = 0
(x - 6)(x - 3) = 0
x = 3 or x = 6
check BOTH in the original equation...you can do that, right ?
What do you THINK about x = 6 ?
Is x = 6 a solution ? Why not ?
[x = 6 is EXTRANEOUS]
then, isolate the radical, square both sides, solve the resulting quadratic equation, check...
√(x - 2) = (4 - x)
(x - 2) = 16 - 8x + x²
x² - 9x + 18 = 0
(x - 6)(x - 3) = 0
x = 3 or x = 6
check BOTH in the original equation...you can do that, right ?
What do you THINK about x = 6 ?
Is x = 6 a solution ? Why not ?
[x = 6 is EXTRANEOUS]
2016-02-11 3:16 pm
Do you mean x + √ (x-2) = 4 ? The root is x = 3.
3 + √ (3-2) = 4
If you really mean x+ √ (x) -2 = 4, the the root is x = 4.
3 + √ (3-2) = 4
If you really mean x+ √ (x) -2 = 4, the the root is x = 4.
2016-02-11 3:25 pm
Case I :
x + √(x - 2) = 4
√(x - 2) = 4 - x
[√(x - 2)]² = [4 - x]²
x - 2 = 16 - 8x + x²
x² - 9x + 18 = 0
(x - 3)(x - 6) = 0
x - 3 = 0 or x - 6 = 0
x = 3 or x = 6 (rejected)
Thus, x = 3
Case II :
x + (√x) - 2 = 4
√x = 6 - x
(√x)² = (6 - x)²
x = 36 - 12x + x²
x² - 13x + 36 = 0
(x - 4)(x - 9) = 0
x - 4 = 0 or x - 9 = 0
x = 4 or x = 9 (rejected)
Thus, x = 4
x + √(x - 2) = 4
√(x - 2) = 4 - x
[√(x - 2)]² = [4 - x]²
x - 2 = 16 - 8x + x²
x² - 9x + 18 = 0
(x - 3)(x - 6) = 0
x - 3 = 0 or x - 6 = 0
x = 3 or x = 6 (rejected)
Thus, x = 3
Case II :
x + (√x) - 2 = 4
√x = 6 - x
(√x)² = (6 - x)²
x = 36 - 12x + x²
x² - 13x + 36 = 0
(x - 4)(x - 9) = 0
x - 4 = 0 or x - 9 = 0
x = 4 or x = 9 (rejected)
Thus, x = 4
2016-02-11 3:17 pm
4-x=sqrt(x-2), square both sides and solve
You get x=3 and x=6, x=3 satisfies.
Hence x=3 is the solution.
You get x=3 and x=6, x=3 satisfies.
Hence x=3 is the solution.
2016-02-11 3:45 pm
substituting √x-2 = t, t>=0 => x = 2+t^2.
rewrite eq. as
2+t^2+t=4, or t^2+t-2=0.
t1=1 => √x-2=1 => x-2=1 => x=3.
while t2<0. Throw it out :)
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Note the function f(x)=x+ √x-2 is increasing on its domain [2; +inf) => the only one solution can exist. But x=3 is an obvious solution.
rewrite eq. as
2+t^2+t=4, or t^2+t-2=0.
t1=1 => √x-2=1 => x-2=1 => x=3.
while t2<0. Throw it out :)
-------------------------------------------------
Note the function f(x)=x+ √x-2 is increasing on its domain [2; +inf) => the only one solution can exist. But x=3 is an obvious solution.
2016-02-11 3:35 pm
x+ √(x−2) = 4
→ √(x−2) = 4−x
Squaring both sides gives
x−2 =16−8x+x²
→x²−9x+18=0
or. x²−6x−3x+18=0
i.e. x(x−6)−3(x−6)=0
→ (x−6)(x−3)=0
Now either (x−6) =0 i.e x= 6 or (x−3)=0 i.e x=3
Check in the equation
When x= 6
LHS=x+√(x−2)=6+√(6−2)=6+√(4)=6+2=8 ≠4≠RHS
Hence x=6 is not a solution
When x= 3
LHS=x+√(x−2)=3+√(3−2)=3+√(1)=3+1=4 =RHS
Hence x=3 is only solution
→ √(x−2) = 4−x
Squaring both sides gives
x−2 =16−8x+x²
→x²−9x+18=0
or. x²−6x−3x+18=0
i.e. x(x−6)−3(x−6)=0
→ (x−6)(x−3)=0
Now either (x−6) =0 i.e x= 6 or (x−3)=0 i.e x=3
Check in the equation
When x= 6
LHS=x+√(x−2)=6+√(6−2)=6+√(4)=6+2=8 ≠4≠RHS
Hence x=6 is not a solution
When x= 3
LHS=x+√(x−2)=3+√(3−2)=3+√(1)=3+1=4 =RHS
Hence x=3 is only solution
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