Project bonus 2: Quadratics with parameters

Given the following equation:

$mx+6-3mx^2=m+4x^2-3x$

1. What values of $m$ make this equation, an equation with 2 roots?

2. Find the values of $m$ so that the equation has only one root as a solution. Calculate the root in each case.

3. Find all the values of $m$ making the equation one without a solution in $\mathbb{R}$ .

4. Find the values of $m$ that give a solution of 4 as one of the roots.

5.Finally, find the values of $m$ making the product of the two roots $-10$ .

Solution
1.To get two roots we have to have a quadratic and have $\Delta>0$
if $a=0$ , this equation will have only one root as solution.
The equation can be written as follows:
$(4+3m)x^2-(3+m)x+m-6=0$
When $a=0 \Rightarrow 4+3m=0$
$3m=-4$
$m=-\frac{4}{3}$
This is the first exception.
Now the discriminant:
$a=4+3m\qquad b=-(3+m) \qquad c=m-6$

(1) $\begin{equation*}\begin{split}\Delta&=b^2-4ac\&=\left[-(3+m)\right]^2-4(4+3m)(m-6)\\&=9+6m+m^2-4(4m-24+3m^2-18m)\\&=9+6m+m^2-4(3m^2-14m-24)\\&=9+6m+m^2-12m^2+56m+96\\&=-11m^2+62m+105\end{split}\end{equation*}$

Question 2:Find the values of $m$ so that the equation has only one root as a solution. Calculate the root in each case.
This is our dream $\Delta=0$
From question 1 we have the 2 solutions
$m=-\frac{15}{11}\; and\; m=7$

First case:
For $m=-\frac{15}{11}$
We plug in $m$ in the equation:
$(4+3m)x^2-(3+m)x+m-6=0$
$(4+3(-\frac{15}{11}))x^2-(3+(-\frac{15}{11}))x+(-\frac{15}{11})-6=0$
$(\frac{44-45}{11})x^2-\frac{33-15}{11}x+frac{-15-66}{11}=0$
$-x^2-18x-81=0$
$\Delta=0$
$x=\frac{18}{-2}$

$x=-9$

When m=7
$(4+3(7))x^2-(3+7)x+7-6=0$
$25x^2-10x+1=0$
$\Delta=0$
$x=\frac{10}{50}$
$x=\frac{1}{5}$

When $m=-\frac{4}{3}$
$a=0$
$(3-\frac{4}{3})x-\frac{4}{3}-6=0$
$\frac{9-4}{3}x-\frac{4}{3}-\frac{18}{3}=0$
$\frac{5}{3}=\frac{22}{3}$

Finally $x=\frac{22}{5}$

Question 3
Find all the values of $m$ making the equation one without a solution in $\mathbb{R}$ .

From our table in question 1:
The solution is $\mathbb{R}-(-\frac{15}{11}, 7)$
Or $\forall m \in (-\infty,-\frac{15}{11})\cup (7, +\infty)$

Question 4: Find the values of $m$ that give a solution of 4 as one of the roots.

One root is:
$x_1=-\frac{3+m+ \sqrt{\Delta}}{2(4+3m)} \Rightarrow -\frac{3+m+ \sqrt{\Delta}}{2(4+3m)}=4$
It yields
$3+m+\sqrt{\Delta}=8(4+3m)$
$\sqrt{\Delta}=32+24m-3-m$
$\Delta=(23m+29)^2$
We get
$-11m^2+62m+105=529m^2+1334m+841$
$540m^2+1272m+736=0$
We simplify
$135m^2+318m+184=0$

(2) $\begin{equation*}\begin{split}\delta&=(318)^2-4 \times 135 \times 184 \\&=101124-99360\&=1764\end{split}\end{equation*}$

Question 5:Finally, find the values of m making the product of the two roots -10.

We have seen in theory that:
$x_1 \cdot x_2=\frac{c}{a}$
In our equation:
$a=(4+3m)$ and $c=m-6$
This gives:
$\frac{m-6}{4+3m}=-10$
$m-6=-40-30m$
$31m=-34$
$m=-\frac{34}{31}$
This value of $m$ is acceptable since it falls within the 2 roots possible solutions interval.

End of project bonus 2

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Project bonus 2: Quadratics with parameters

Project bonus 2: Quadratics with parameters

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