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Hey there
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The correct answer is:
f(x+5)=(x+5)^2+3(x+5)-10

f(x+5)=x^2+10x+25+3x+15-10

f(x+5)=x^2+13x+30  and f(x+5)=x^2+kx+30 so

k=13

Now factor x^2+13x+30

Find j and k such that jk=ac=30 and j+k+b=13 so j and k are 10 and 3 so

(x+3)(x+10)

So the two zeros occur when x=-3 and -10 the smallest of which is:

x=-10

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Hope this helps you
susanwiederspan
The zeros of [tex]x^2+3x+30[/tex] can be obtained by the quadratic formula (we use this method because the function is not factorable and has no real solutions, so it is easiest).

For a general quadratic [tex]ax^2+bx+c=0[/tex],

[tex]x= \dfrac{-b \pm \sqrt{b^2-4ac} }{2a} [/tex].

For our function, a = 1, b = 3, and c = 30.

We substitute and solve for x:

[tex] =\dfrac{-(3) \pm \sqrt{(3)^2-4(1)(30)} }{2(1)}= \dfrac{-3 \pm \sqrt{9-120} }{2}\\\\\\= \dfrac{-3 \pm \sqrt{-111} }{2}=\dfrac{-3 \pm i\sqrt{111} }{2}[/tex]

[tex]x=\dfrac{-3+i \sqrt{111}}{2} \ \\\\\\\ x = \dfrac{-3-i \sqrt{111}}{2} \ [/tex]

The modulus of each complex solution is about 5.477, so neither solution is bigger or smaller.

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