Question

17. When √2z+1+ √3z+4 = 7 the value of z is given by​

Answer 1

Step-by-step explanation:

(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

Answer 2

Step-by-step explanation:

hey mate here is your answer:

(√2z+1+√3z+4) = 7

(√2z+√3z+5) = 7

z(√2+√3)=7-5

z(√2+√3)= 2

.:squaring on both sides.

[z(√2+√3)]^2 = 2^2

z^2(2+3+2√6)=4