How do we know that the quadratic $3y^2-y-12$ has real root?

Solution 1:

(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 $$3y^2-13y+2.$$

Solution 2:

Ways to tell if a quadractic has a root.

1. A quadratic has a minimum or maximum depending on the value of $a$ in $ay^2$. If $a > 0$, there is a minimum, if $a < 0$ there is a maximum. Since, your $a = 3, a > 0$ in $3y^2-y-12$, its a minimum. Taking $\frac{d}{dy} = 0, 6y - 1 = 0, y = \frac{1}{6}$. $f(\frac{1}{6}) < 0$, (below the horizontal-axis). Since its absolute maximum is $y = \infty$, it must pass through the horizontal-axis ($f(y) = 0$) twice.

2. You can take its discriminant, $b^2 - 4ac$ from $ay^2 + by + c$. If $b^2 - 4ac > 0$, it has $2$ real roots, or if $b^2 - 4ac = 0$, it has $1$ root, or if $b^2 - 4ac < 0$, it has imaginary roots (no real roots).

3. Use the quadratic formula (this is built on the discriminant). $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

4. Factor your quadratics. With yours you need to complete the square first. $3(y-\frac{1}{6})^2-\frac{145}{12}.$ The factor the $-\frac{145}{12}$ in.

$\to 3(y-\frac{1}{6})^2-\frac{145}{12}$

$\to 3((y-\frac{1}{6})^2-\frac{145}{36})$

$\to 3(y-\frac{1}{6} + \frac{\sqrt{145}}{6})(y-\frac{1}{6} - \frac{\sqrt{145}}{6})$

Hence your roots are $\frac{1}{6} \pm \frac{\sqrt{145}}{6}$. Compare this with ... $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ ... $\frac{1 \pm \sqrt{1 - (4)(3)(-12)}}{6}$ ... $\frac{1 \pm \sqrt{145}}{6}$ ... $\frac{1}{6} \pm \frac{\sqrt{145}}{6}$

Solution 3:

You can use various methods. Your method, but, is useful for finding rational roots.

The quadratic is irreducible over rationals.

Let $f(y)=3y^2-y-12$. If $f(y)$ is reducible over rationals, then one of these should be $0$: $f(\pm1), f(\pm2), f(\pm3), f(\pm4), f(\pm6), f(\pm12), f(\pm\frac13), f(\pm\frac23), f(\pm\frac43)$.

But they are all non-zero. So $f(y)$ is not reducible on rationals, by rational root theorem.

While using the root formula may be the surest way, but if the exact root value is not required, solving it by using IVP is one method.

$f$ is polynomial function, so it is continuous, indeed. So, you can use Intermediate Value Theorem(IVP).

$f(2)=-2, f(3)=12$. So by IVP, there exists $2<t<3$ such that $f(t)=0$, because $f(2)<0<f(3)$.

$y=t$ is sure a real root of $f(y)$.

Solution 4:

Using Vieta theorem, $r+s=\frac{1}{3}, r\cdot s=-4$.
Thus, $(r+2)(s+2)=rs+2(r+s)+4=-4+\frac{2}{3}+4=\frac{2}{3}$
and $(r+2)+(s+2)=r+s+4=\frac{13}{3}$
Applying Vieta in reverse, we can get a solution for c): $3y^2-13y+2=0$

To show that there is a real root, rewrite the equation as $3(x-\frac{1}{6})^2=12+\frac{1}{12}$ 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.

Solution 5:

The only requirement for a real root is $\space b^2-4ac\ge 0.\quad$ In this case, $\quad -4ac=-\big(4(3)(-12)\big)=+144\quad$ so $b^2+144>0 \space$ and $\space y=\dfrac{-b\pm\sqrt{b^2+144}}{6}\space$ which is real.

Some integer solutions (perhaps the only ones) for $\space (b,y_1,y_2)\space$ are $$ (5,4,-9)\quad (9,3,-12)\quad (16,2,-18)\quad (35,1,-36)\quad $$ and, it appears that $y_2=-(b+y_1).\quad$ Whether or not this can be generalized to all real values of $b$ and $y_1,\space$ the equation will always have real roots.