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

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Answer 1 · 2018-06-21T10:30:16

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

$x = 0, x = - 22$ $x=0, x=-22$

Given: $y = 0.05 x^{2} + 1.1 x + 0$ $y=0.05x^2+1.1x +0$

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

Set $y = 0 = 0.05 x^{2} + 1.1 x$ $y=0=0.05x^2+1.1x$

Factor out $x \to 0 = x (0.05 x + 1.1)$ $x -> 0=x(0.05x+1.1)$

Set $0.05 x + 1.1 = 0 \Rightarrow 0.05 x = - 1.1$ $0.05x+1.1=0 =>0.05x=-1.1$

$x = - \frac{1.1}{0.05} = - \frac{110}{5} = - 22$ $x=-1.1/0.05 = -110/5 = -22$

$x = 0 and - 22$ $x=0 and -22$

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

Using 0 as a place keeper

$y = a x^{x} + b x + c \to at y=0 we have x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $y=ax^x+bx+c ->" at y=0 we have "x=(-b+-sqrt(b^2-4ac))/(2a)$

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

$x = \frac{- 1.1 \pm \sqrt{{(1.1)}^{2} - 4 (0.05) (0)}}{2 (0.05)}$ $x=(-1.1+-sqrt((1.1)^2-4(0.05)(0)))/(2(0.05))$

Anything times zero is zero so we have:

$x = \frac{- 1.1 \pm \sqrt{{(1.1)}^{2}}}{2 (0.05)}$ $x=(-1.1+-sqrt((1.1)^2))/(2(0.05))$

$x = - \frac{1.1}{0.1} \pm \frac{1.1}{0.1}$ $x=-1.1/0.1+-1.1/0.1$

$x = - 11 \pm 11$ $x=-11+-11$

$x = 0, x = - 22$ $x=0, x=-22$

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

Tony B

How do you find the solution to the quadratic equation Y=0.05x2+1.1xY=0.05x^2 + 1.1x?