Consider the quadratic $3y^2-y-12$.
(a) Notice the quadratic cannot be factored into the product of two binomials with integer cofficients. Does this mean that the quadratic does not have any real roots?
(b) If the answer to part (a) is "no", then explain how we know that the quadratic does have real roots.
(c) Suppose the quadratic has roots $y=r$ and $y=s$. Find a quadratic with roots $r+2$ and $s+2$.
$(Ay+B)(Cy+D) = ACy^2+(AD+BC)y+BD = 3y^2-y-12$
$AC=3\\
BD=-12\\
AD + BC = -1$
A hint is given to evaluate the quadratic for $y=0$ and $y=3$. However I didn't find it useful. How was this hint suppose to be helpful?
$3(0)^2-0-12=-12 \\
3(3)^2-3-12=12$
When trying to factor using integer cofficients, I'm unsuccesful in producing the quadratic i.e.
$(A,C) = (1,3)$ or $(3,1)$
$(B,D) = (-4,3)$ or $(4,-3)$ or $(-6,2)$ or $(6,-2)$ or $(-12,1)$ or $(12,-1)$
I figure then I must use rational numbers for the cofficients and/or constants but this is proving a difficult guessing game.
I'm still mulling over part (c). Any guidance here would be helpful.
Best Answer
(a) This does not mean the quadratic has no real roots. A clearer example of this is $$y^2-2=(y+\sqrt{2})(y-\sqrt{2}).$$ (b) One way to know is to observe that plugging in $y=0$ and $y=3$ yields $$3\cdot0^2-0-12=-12,$$ $$3\cdot3^2-3-12=12.$$ so somewhere between $0$ and $3$ the quadratic must equal $0$.
Another way to know is by computing the discriminant, which is $$\Delta=b^2-4ac=(-1)^2-4\cdot3\cdot(-12)=145.$$ The quadratic has a real root because the discriminant is nonnegative.
(c) If $r$ and $s$ are roots of $$3y^2-y-12,$$ then it follows that $r+2$ and $s+2$ are roots of $$3(y-2)^2-(y-2)-12,$$ which by a little bit of algebra simplifies to $$3y^2-13y+2.$$