functions Q2

1. It is given that f(x)=(x-5)(x-6)-k^2+20.

a)If f(k)=0,find the value of k.
f(k)=(k-5)(k-6)-k^2+20=0
k^2-6k-5k+30-k^2+20=0
-11k+50=0
k=-50/11

b)Hence,solve for x if f(x+1)=0
f(x+1)=(x+1-5)(x+1-6)-(-50/11)^2+20=0
(x-4)(x-5)-(2500/121)+20=0
x^2-5x-4x+20-(2500/121)+20=0
x^2-9x+[40-(2500/121)]=0
x=5.45 or x=3.55 (to 3 sig fig) [By 計數機]

1(a):::
f(x)=(x-5)(x-6)-k^2+20
f(k)=(k-5)(k-6)-k^2+20
0=k^2-11k+30-k^2+20
11k=50
k=50/11

1(b):::
f(x+1)=[(x+1)-5][(x+1)-6]-(50/11)^2+20
0=(x-4)(x-5)-80/121
x^2-9x+20-80/121=0
x^2-9x+2340/121=0
x=60/11 or x=39/11
x=5.455 or x=3.545 (cor. to 3 sig. fig.)