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

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Answer 1 · 2018-06-21T10:30:16

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

$x=0, x=-22$

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: $y=x(x+22)$

Tony B

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