treslagosnv

2022-02-01

How do we know that the quadratic $3{y}^{2}-y-12$ has real root?
(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 . Find a quadratic with roots .

Amari Larsen

Expert

(a) This does not mean the quadratic has no real roots. A clearer example of this is
${y}^{2}-2=\left(y+\sqrt{2}\right)\left(y-\sqrt{2}\right)$.
(b) One way to know is to observe that plugging in yields
$3\cdot {0}^{2}-0-12=-12$,
$3\cdot {3}^{2}-3-12=12$.
so somewhere between 0 and 3 the quadratic must equal 0.
Another way to know is by computing the discriminant, which is
$\mathrm{\Delta }={b}^{2}-4ac={\left(-1\right)}^{2}-4\cdot 3\cdot \left(-12\right)=145.$
The quadratic has a real root because the discriminant is nonnegative.
(c) If r and s are roots of
$3{y}^{2}-y-12,$
then it follows that are roots of
$3{\left(y-2\right)}^{2}-\left(y-2\right)-12$
which by a little bit of algebra simplifies to
$3{y}^{2}-13y+2$.

Nevaeh Jensen

Expert

Using Vieta theorem, $r+s=\frac{1}{3},r\cdot s=-4$.
Thus, $\left(r+2\right)\left(s+2\right)=rs+2\left(r+s\right)+4=-4+\frac{2}{3}+4=\frac{2}{3}$
and $\left(r+2\right)+\left(s+2\right)=r+s+4=\frac{13}{3}$
Applying Vieta in reverse, we can get a solution for c):
$3{y}^{2}-13y+2=0$
To show that there is a real root, rewrite the equation as $3{\left(x-\frac{1}{6}\right)}^{2}=12+\frac{1}{12}$ and use that any non-negative real can be written as a square of some real. Or you can use the hint provided and apply Intermediate Value Theorem.

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