ANSWER
[tex]x = 2 \: \: or \: \: x = 15[/tex]
Or
[tex]x = - 2 \: \: or \: \: x = - 15[/tex]
EXPLANATION
The given polynomial is
[tex] f(x) = {x}^{2} + kx + 30[/tex]
where a=1,b=k, c=30
Let the zeroes of this polynomial be m and n.
Then the sum of roots is
[tex]m + n = - \frac{b}{a} = -k [/tex]
and the product of roots is
[tex]mn = \frac{c}{a} = 30[/tex]
The square difference of the zeroes is given by the expression.
[tex]( {m - n})^{2} = {(m + n)}^{2} - 4mn [/tex]
From the question, this difference is 169.
This implies that:
[tex]( { - k)}^{2} - 4(30) = 169[/tex]
[tex]{ k}^{2} -120= 169[/tex]
[tex] k^{2} = 289[/tex]
[tex] k= \pm \sqrt{289} [/tex]
[tex]k= \pm17[/tex]
We substitute the values of k into the equation and solve for x.
[tex]f(x) = {x}^{2} \pm17x + 30[/tex]
[tex]f(x) = (x \pm2)(x \pm 15)[/tex]
The zeroes are given by;
[tex] (x \pm2)(x \pm 15) = 0[/tex]
[tex]x = \pm2 \: \: or \: \: x = \pm 15[/tex]
