Prove that all the solutions of the equation (x + 1)^1/2 = 5 - x are consistent

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hala718 eNotes educator| Certified Educator

(x+1)^1/2 = 5-x

First (x+1) must be greater that 0

==> x+1 > 0

==> x > -1

And 5-x > 0 ==> x < 5

==> x = [-1,5]

Now let us solve:

Square both sides:

==> (x+1) = (5-x)^2

==> x+1 = 25 - 10x + x^2

==>x^2 -11x + 24 = 0

==> (x-8)(x-3) = 0

==> x1= 3

==> x2= 8

we note that x1 belongs to the interval [-1,5] ,  -1< 3< 5

But x2 is not consistent since 8 > 5

Then the only solution is:

x = 3

 

neela | Student

To find the solutions and whether they are consistent.

(x+1)^(1/2) = 5-x.

We square both sides:

x+1 = (5-x)^2

x+1 = 25-10x+x^2

0 = 25-1 -10x-x +x^2.

x^2-11x+24 = 0

(x-8)(x-3) = 0.

x = 8 and x = 3 are the solutions.

Substite x= 8 and x = 3 in (1+x)^(/2) = 5-x.

x= 8:

LHS (1+8)^(1/2) = 3 or -3. RHS : 5-8 = 3. Verifies for x= 8.

x =3:

LHS: (1+3)^1/2 = 2or -2. RHS:  5-3 = 2 . Verifies for x = 3.

So the  all  soltions are consistent.

 

giorgiana1976 | Student

Before finding the solutions of the equation, we'll have to impose the conditions of existence of the square root.

x + 1>=0

x>=-1

and

5 - x>=0

-x>=-5

x=<5

The solutions of the equation have to belong to the interval of valid values [-1 , 5].

Now, we'll solve the equation. We'll substitute the exponent (1/2) by the term square root = sqrt.

sqrt (x + 1) = 5 - x

We'll raise to square both sides:

[sqrt (x + 1)]^2 = (5 - x)^2

We'll eliminate the square root from the left side and we'll expand the square from the right side:

x + 1 = 25 - 10x + x^2

We'll subtract x+1 both side, we'll combine like terms and we'll use the symmetric property:

x^2 - 11x + 24 = 0

We'll apply the quadratic formula:

x1 = [11 + sqrt(121 - 96)]/2

x1 = (11+5)/2

x1 = 8

x2 = (11-5)/2

x2 = 3

Since the first value for x1 = 8, doesn't belong to the interval of admissible values [-1 , 5], we'll accept as valid solution only x2 = 3.

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