Quadratic expression and equations from additional mathematics

Question

Show that the line $x+ y = q$ will intersect the curve $x^2-2x+2y^2=3$ in two distinct points where $q^2<2q+5$

Answer 1

We want the curve to be in simpler terms. So we will sub in for $y$: $$x+y=q$$ $$y=q-x$$ Plugging into curve equation: $$x^2-2x+2(q-x)^2=3$$ $$x^2-2x+2(q^2-2xq+x^2)=3$$ $$x^2-2x+2q^2-4xq+2x^2=3$$ $$3x^2-(2+4q)x+(2q^2-3)=0$$

Use the discriminant to find how many intersections there will be. For 2 real intersections: $${b^2-4ac}>0$$ Look at the curve equation to find the coefficients. $${[−(2+4q)]^2-4(3)(2q^2-3)}>0$$ $${(4+16q+16q^2)-12(2q^2-3)}>0$$ $${4+16q+16q^2-24q^2+36}>0$$ $${-8q^2+16q+40}>0$$ Which simplifies to: $$-q^2+2q+5>0$$ Look at the form you need it in. Just move over. $$q^2<2q+5$$

Answer 2

Hint: Plugging $$y=q-x$$ in the equation above we get $$x^2-2x+2(q^2-2qx+x^2)-3=0$$ so we have to solve $$3x^2-2x(2q+1)+2q^2-3=0$$ Can you proceed? The discriminant must be $$(2q+1)^2>6q^2-9$$ this is $$0>q^2-2q-5$$