2題MATHS

1.The equation x^2-bx-24=0 can be solved using the factor method.Suggest two possible values for b.

∵-24
=-2×12 or
=-3×8 or
=-4×6 or
=-12×2 or
=-8×3 or
=-6×4
possible values for b can be:
2,5,10,-2,-5,-10
-2+12=10
-3+8=5
-4+6=2
-12+2=-10,
-8+3=-5
-6+4=-2


2.If the roots of the equation x^2-18x+c=0 are two distinct integers of the same sign,suggest two possible values for c.

Then x^2 - 18x+c can be factorize into the form (x + p) (x + q)
but p and q is -18 and they are of the same sign.
Terefore the correct form should be (x - p) (x - q)

Thus the only possible values for c are
17,32,45,56,65,72,77,80,81

-1 x -17 = 17
-2 x -16 = 32
-3 x -15 = 45
-4 x -14 = 56
-5 x -13 = 65
-6 x -12 = 72
-7 x -11 = 77
-8 x -10 = 80
-9 x -9 = 81

1.-5,,,2

2,,b^2+4ac=0
(-18)^2-4(1)(c)=0
324-4c=0
c=81