Q1: Let's suppose that the 2 numbers are "a" and "b".
So a+b=7 and a^2 + b^2=109
But a^2 + b^2=(a+b)^2-2ab
So, a^2 + b^2=109 and a+b=7, that means that after subtitution: 109=7^2-2ab
109-49=-2ab
60=-2ab
ab=-30 and a+b=7+>a=7-b
We'll substitute a=7-b, into the product ab=-30
(7-b)*b=-30
7b-b^2-30=0
b^2-7b-30=0
b1=[7+sqrt(49-120)]/2=(7+13)/2=20/2=10
b2=[7-sqrt(49-120)]/2=(7-13)/2=-6/2=-3
So a1=7-b1=7-10=-3
a2=7-b2=7-(-3)=7+3=10
The system formed with the expression a+b=7, a^2+b^2=109 is known as symmetrical sustem, where the solution of the 2 unknown could be changed between them without affecting the system solution.
Q2: h=30t-5t^2
a) When h=0, the expression above becomes 0=30t-5t^2
We'll multiply the equation formed with the value (-1) and we'll re-write the equation.:
5t^2-30t =0
We've noted t as common factor and we'll use this in this way:
t(5t-30)=0, knowing that a product of 2 factors which is equal to 0, one of the 2 factors is 0.
So, t=0, or
5t-30=0
t-6=0
t=6
b) When h=25, 30t-5t^2 = 25
30t-5t^2- 25=0
We'll divide the expression with 5
6t - 5t^2 - 5=0
We'll multiply the expression with (-1)
5t^2-6t+5=0,
As we can see, the expression above is always bigger than the value 0, and never 0, for any value of t. The estimation was made based on fact that delta=(-6)^2+4*5*5<0.
In terms of graphics, the expression 5t^2-6t+5=0, represents a curve with parabolic shape (due to the second grade of the equation), this curve being placed all the above of the ox axis, never intersecting it, for any values of t!
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