z^2-14z+48=0

Question

solution for the quadratic equation. please show steps. could it be -6,8  Thank you

Answer 1

Hi Roseland,

 

z²-14z+48=0

 

z²-6z-8z+48=0

 

z(z-6)-8(z-6)=0

 

(z-8)(z-6)=0

 

z-8=0     

 

z=8

 

z-6=0

 

z=6

Answer 2

In the quadratic roots formula the term D= √(b²-4*a*c) determines if the roots are real or imaginary in this case b=-14, a=1 and c=48 so D=2 so the roots are real and are (14+-2)/2=8,6.

 

This quadratic factors into (x-6)(x-8) from which you can see the roots are 8 and 6.

 

Jim

Answer 3

z^2 - 14z + 48 = 0 is in Standard Form.

 

(z - h)^2 + k = 0 would be Vertex Form if we knew h and k.

 

Expand:

 

z^2 - 2hz + h^2 + k = 0 is Vertex Form.

z^2 - 14z + 48 = 0 is Standard Form.

 

2h = 14, h = 7

h^2 + k = 48, k = 48 - 7^2 = 48 - 49 = -1

 

(z - 7)^2 - 1 = 0 is Vertex Form; solve for z:

 

(z - 7)^2 = 1

 

|z - 7| = 1

 

z - 7 = ±1

 

z = 7 ± 1 = 6, 8

 

check:

 

(6)^2 - 14(6) + 48 =? 0

36 - 84 + 48 =? 0

84 - 84 = 0 √

 

(8)^2 - 14(8) + 48 =? 0

64 - 112 + 48 =? 0

112 - 112 = 0 √