Expand the following:
(5 a + b/5)^2
(5 a + b/5) (5 a + b/5) = (5 a) (5 a) + (5 a) (b/5) + (b/5) (5 a) + (b/5) (b/5):
5×5 a a + (5 a b)/5 + (5 b a)/5 + (b b)/(5×5)
(5 a b)/5 = 5/5×a b = a b:
5×5 a a + a b + (5 b a)/5 + (b b)/(5×5)
(b×5 a)/5 = 5/5×b a = b a:
5×5 a a + a b + b a + (b b)/(5×5)
Combine powers. (b b)/(5×5) = (b^(1 + 1))/(5×5):
5×5 a a + a b + b a + (b^(1 + 1))/(5×5)
1 + 1 = 2:
5×5 a a + a b + b a + (b^2/5)/5
5 a×5 a = 5×5 a^2:
5×5 a^2 + a b + b a + (b^2/5)/5
5×5 = 25:
Answer: 25 a^2 + a b + b a + (b^2/5)/5
When a quadratic equation ax^2+bx+c has a double root, the discriminant,
D=b^2-4ac=0
Here
a=2,
b=b,
c=18
and
D=b^2-4ac=b^2-4*2*18=0
solve for b
b^2-144=0
=> b= ± sqrt(144)= ± 12
So in order that the given equation has double roots, the possible values of b are ± 12.
Answer: 3.5
Step-by-step explanation: divide 49 by 14
Fourth test mark should be 86
85 + 82 + 75 + 86 = 328
328 divide by 4 (as there are four tests) = 82
