Part (b)
We use the result of part (a) and plug in (x,y) = (0,0). This is directly from the initial condition y(0) = 0.
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This means,
is the solution with the initial condition y(0) = 0.
3x^2+11x-4=10x-1
3x^2+11x-10x-4+1=0
3x^2+x-3=0
Δ=1^1-4*3*(-3)=1+36=37
x1=(-1+V37)/6
x2=( -1-V37)/6
Answer:
Step-by-step explanation:
-7z/2 - 2 + z -3
=-5 - 5z/2
ANSWER
EXPLANATION
The quadratic equation is:
Group variable terms:
Add the square of half, the coefficient of y to both sides.
The LHS us now a perfect square trinomial:
Take square root:
The first choice is correct.