(2z+1)^1/2+(3z+4)^1/2 = 7, what is the value of z?

Question

Choices:-

A 1

B 2

C 3

D 4

 

Correct answer is 4. How?

Answer 1

Move one of the terms to the right side of the equation.

 

(2z + 1)^1/2 = 7 - (3z + 4)^1/2

 

 

Square both sides of the equation.

 

2z + 1 = 49 - 14√(3z + 4) + 3z + 4

 

2z + 1 = 53 + 3z - 14√(3z + 4) 

 

-z - 52 = -14√(3z + 4) 

 

 

Square both sides of the equation.

 

z² + 104z + 2704 = 196(3z + 4)

 

z² - 484z + 1920 = 0

 

You now have a quadratic equation.  Use the quadratic formula to solve for z.  See what you get.

 

Answer 2

Hi Rahul:

 

This is a tedious problem but works out as below:

 

(2z+1)^1/2 + (3z+4)^1/2 = 7

 

Square both sides

 

[(2z+1)^1/2 + (3z+4)^1/2]² = 7²

 

Remember:  (a+b)² = a² + 2ab + b²

a = 2z+1   ...  b = 3z+4

and

(√a)(√b) = √(ab)

 

2z+1 + 2[(2z+1)(3z+4)]^1/2 + 3z + 4 = 49

 

5z + 5 + 2(6z² + 11z + 4)^1/2 = 49

 

2(6z² + 11z + 4)^1/2 = 44 - 5z

 

square both sides

 

4(6z² + 11 + 4) = (44-5z)²

 

24z² + 44z + 16 = 1936 - 440z + 25z²

 

Move everything to right side

 

z² - 484z + 1920 = 0

 

We now have a quadratic equation we can use the quadratic formula on

 

z = [484 ±√((-484)²-4(1)(1920))]/2(1)

 

z = (484 ±√226576)/2

 

z = (484 ± 476)/2

 

z = 8/2 = 4   or   z = 960/2 = 480

 

Do you see?