x = ? , 5 , 9 , 4
y= 0, ? , 1 , ?
To find the missing x value, replace the matching value of y (0) in the equation and solve for x:
0 = x^2-12x +36
Apply the quadratic formula
[tex]\frac{-b\pm\sqrt[]{b^2-4\cdot A\cdot c}}{2\cdot a}=\frac{12\pm\sqrt[]{(-12)^2-4\cdot1\cdot36}}{2\cdot1}[/tex][tex]\frac{12\pm\sqrt[]{144-144}}{2}=\frac{12}{2}=6[/tex]For x = 5:
y= (5)^2-12 (5) +36 = 25-60+36 = 1
For y=1
1 =x^2-12x+36
0 = x^2-12x+36-1
0= x^2-12x+35
[tex]\frac{12\pm\sqrt[]{(12)^2-4\cdot1\cdot35}}{2\cdot1}=\frac{12\pm\sqrt[]{144-140}}{2}=\frac{12\pm2}{2}=\frac{14}{2}=7\text{ }[/tex]x =7
For x=9
y= (9)^2-12 (9)+36 = 81-108+36=9
For x=4
y= (4)^2-12(4)+35 = 16-48+36=4