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