find the sum and product of the roots of 8×2+4×+5=0
5 x n = (3 x 16) - 8 5 x n = 48 - 8 5 x n = 40 n = 40/5 ie 8
It is 8*(-4)N which is -32N
n - 6m = 8 n = 6m + 8
-15
mn = 8 ( Product) m + n = 6 (Sum) Hence m = 6 - n Substitute (6 -n)n = 8 Multiply out the brackets 6n - n^(2) = 8 n^(2) - 6n + 8 = 0 It is now in quadratic form and will factor Hence ((n - 2)(n - 4) = 0 n = 2 & n = 4 So '2' & '4' are the two numbers. Verification 2 + 4 = 6 2 x 4 = 8
(9n - 8) should do it.
Is 4 cause it is equivalent to 2 n a multiple of 8 for example n./. 8 so that the answer
Let the consecutive integers be 'n' & 'n+1'. Their product is n(n+1) Their sum is n + n+ 1 = 2n + 1 Eight timres the sum is 8(2n+1) Their product being 8 less that their sum is n(n+1) = 8(2n+1) - 8 Hence n^(2) + n = 16n + 8 - 8 n^(2) + n = 16n n^(2) = 15n ( Cancel down by 'n'). n = 15 n+1 = 16 Hence '15' & '16' are the two numbers.
The product of 15 and a number (when the number is n) is 15n. Subtract 8 from this and we have 15n-8
find the sum and product of the roots of 8×2+4×+5=0
In algebra, 8n means to multiply n and 8. (8xN)
8n+6
5 x n = (3 x 16) - 8 5 x n = 48 - 8 5 x n = 40 n = 40/5 ie 8
8-x 8-x
It is 8*(-4)N which is -32N
8-n