Question

How do you find the solution to the quadratic equation `Y=0.05x^2 + 1.1x`?

Answer 1

Solving fo `x` when `y=0 -> x" intercepts"`

`x=0, x=-22`

Explanation:

Given: `y=0.05x^2+1.1x +0`

`color(blue)("The more efficient - it is a matter of spotting it.")`

Set `y=0=0.05x^2+1.1x`

Factor out `x -> 0=x(0.05x+1.1)`

Set `0.05x+1.1=0 =>0.05x=-1.1`

`x=-1.1/0.05 = -110/5 = -22`

`x=0 and -22`

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`color(blue)("The inefficient way: -Using the formula")`

Using 0 as a place keeper

`y=ax^x+bx+c ->" at y=0 we have "x=(-b+-sqrt(b^2-4ac))/(2a)`

In this case `a=0.05; b=1.1; c=0`

`x=(-1.1+-sqrt((1.1)^2-4(0.05)(0)))/(2(0.05))`

Anything times zero is zero so we have:

`x=(-1.1+-sqrt((1.1)^2))/(2(0.05))`

`x=-1.1/0.1+-1.1/0.1`

`x=-11+-11`

`x=0, x=-22`

Thus we have in factored form: `y=x(x+22)`