In below problem I've been trying to figure out how simply letting $y=\dfrac{x}{x+10}$ gives the quadratic with roots scaled by $\dfrac{1}{\alpha+10}$. I'm a bit clueless why it works.
My thoughts :
- To get a quadratic $g(x)$ whose roots are $k$ times bigger than $f(x)$, we can simply replace $x$ by $x/k$. This works because $f(\alpha) = 0 \implies g(k\alpha) = 0$, where $g(x) = f(x/k).$
- To get a quadratic $g(x)$ whose roots are $h$ bigger than $f(x)$, we can simply replace $x$ by $x-h$. This works because $f(\alpha) = 0 \implies g(\alpha+h) = 0$, where $g(x) = f(x-h).$
The roots of $$x^2-13x-2=0$$ satisfy $$a+b =13, ab=-2$$
The transformed equation is $$(x-\frac {a}{a-10})(x-\frac {b}{b-10}) $$
Upon simplification we get $$(ab-10(a+b)+100)x^2 -(2ab-10(a+b))x+ab=0$$
Substitution for a+b and ab gives us $$-32x^2+134x-2=0$$
Thus the correct answer is $$-32x^2+134x-2=0$$