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

Question

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.

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

Answer 2

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 $f(y)$.

Answer 3

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.

Answer 4

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 ... $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ ... $\frac{1 \pm \sqrt{1 - (4)(3)(-12)}}{6}$ ... $\frac{1 \pm \sqrt{145}}{6}$ ... $\frac{1}{6} \pm \frac{\sqrt{145}}{6}$

Answer 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.

Answer 6

For (b), still another way to determine whether the quadratic polynomial has real roots is to consider the graph of its curve. The parabola corresponding to $ \ y \ = \ a·(x-h)^2 \ $ has its vertex on the $ \ x-$axis, so the function equals zero only for $ \ x = h \ \ ; $ the function is then $ \ ax^2 - 2ahx + ah^2 \ \ . $ If we "shift" the parabola "downward" by adding a negative number to the function (thus, $ \ a·(x-h)^2 - k \ $ ), the vertex will be "below" the $ \ x-$axis and the parabola will intersect that axis twice, corresponding to the two zeroes of the quadratic polynomial. So if a quadratic polynomial has a constant term less than or equal to $ \ ah^2 \ \ , $ it has one ( constant term $ = ah^2 \ $ ) or two ( constant term $ < ah^2 \ $ ) real roots.

[This is essentially the analytic geometry behind Vasya's answer (and a couple of the comments under the OP) for this part; this test is based on "completing the square", but uses only minimal calculation. It also contains the information given by the discriminant, but is less work to compute.]

The quadratic in question is $ \ 3y^2 - y - 12 \ \ , $ so we have $$ a = 3 \ \ , \ -2ah = -1 \ \ \Rightarrow \ \ h = \frac16 \ \ \Rightarrow \ \ ah^2 = \ 3·\frac{1}{6^2} = \frac{1}{12} \ \ . $$ The constant term $ \ -12 \ $ is definitely smaller than $ \ ah^2 \ \ , $ so this quadratic polynomial has two real roots.