If the equation $|x^2+4x+3|-mx+2m=0$ has exactly three solutions then find value of m.

Question

Problem :

If the equation $|x^2+4x+3|-mx+2m=0$ has exactly three solutions then find value of $m$.

My Approach:

$|x^2+4x+3|-mx+2m=0$

Case I : $x^2+4x+3-mx+2m=0$

$\Rightarrow x^2+ x (4-m) + 3+2m=0 $

Discriminant of above qudratic is

$D = (4-m)^2 -4(3+2m) \geq 0$

$D = 16+m^2-8m-12-8m$ Solving for $m$ we get the values $-8 \pm 2\sqrt{15}$

Case II :

Similarly solving for the given equation taking negative sign of modulus we get the solution

for $m =$$8 \pm 2\sqrt{15}$

Can we take all the values of m to satisfy the given condition of the problem , please suggest which value of m should be neglected in this. Thanks.

Answer 1

The "three solutions" referred to are three distinct values of $x$ that make the equation true. As you will see later, one of those values must be a repeated root, so arguably there are four roots, but let that go for the moment...

This equation can be restated as $|x^2+4x+3|=m(x-2)$. Writing it this way, it becomes clear that we are looking for points of intersection of the line $y=m(x-2)$ and the curve $y=|x^2+4x+3|$.

The curve is large and positive for large negative values of $x$, decreases to 0 at $x=-3$, rises to a maximum at $x=-2$ and falls to ) at $x=-1$ before rising again to become large and positive as $x$ becomes large and positive. This pattern can be described as a valley with a hill on the valley floor.

The line must pass through the point $(2,0)$, but its gradient depends on $m$. Lines with a positive gradient will never intersect the curve, so the gradient must be negative.

There are three possible types of intersection:

1) The line does not intersect the "hill", so intersects the curve exactly twice (at the valley walls). This means the equation would have exactly two roots.

2) The line intersects the "hill" twice as well as the valley walls. This means the equation would have exactly four roots.

3) The line is a tangent to the "hill" as well as intersecting the valley walls. This means the equation would have exactly three roots.

It is the third scenario that you need to find.

This is the "Type 2" that you have referred to in your question.

The equation is $-x^2-4x-3=m(x-2)$

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

$-x^2-4x-mx-3+2m=0$

$x^2+(4+m)x+(3-2m)=0$

Repeated root means $(4+m)^2-4(3-2m)=0$

Like you I get $m=-8\pm2\sqrt{15}$.

The very steep root can be discounted because it will not work.

So $m=-8+2\sqrt{15}$.

Answer 2

$$m(x-2)=|(x+3)(x+1)|\ge0$$

If $m=0,$ there are two real solutions

Else $m(x-2)=|(x+3)(x+1)|=0$ has no solution

So, $$m(x-2)=|(x+3)(x+1)|>0$$

Now $|(x+3)(x+1)|=-(x+3)(x+1)$ if $-3\le x\le-1$

$=+(x+3)(x+1)$ otherwise

If $m>0,x-2>0\iff x>2\implies m(x-2)=x^2+4x+3$ which has exactly two solutions

If $m<0,x-2<0\iff x<2$

If $-1<x<2$ or if $x<-3;$ $m(x-2)=(x+3)(x+1)\ \ \ \ (1)$

If $-3\le x\le-1,m(x-2)=-(x+3)(x+1)\ \ \ \ (2)$

We need the discriminant of $(1)$ or $(2)$ to be zero honoring the range.

Answer 3

the value of $m$ you are looking for is the slope of the tangent line to the graph $y = -x^2 - 4x _ 3, -3 < x < 1.$ it has a tangent at the $x$-coordinate $2 - \sqrt {15}$ with a slope of $-2a - 4 = -8 + 2\sqrt{15}.$ any line with a bigger slope cuts at two points and any with a positive and smaller slope will cut at four points and if the slope is zero, then it cuts at two pints $x = -1, x = -3.$

Answer 4

The parabola $x^2 + 4x + 3 = (x+3)(x+1)$ is upward facing and has roots at $x = -3, x = -1$, so it is negative in $(-3, -1)$. Break it into cases. One of these cases has two solutions, the other has exactly one (i.e. discriminant = 0).

Case 1: $-3 \leq x \leq -1$. The equation is $-x^2 - 4x - 3 - mx + 2m = 0$ so $x^2 + (4+m)x + (3-2m) = 0$. Discriminant is $\sqrt{(m+4)^2 - 4(3-2m)}$.

Case 2: The other case. The equation is $x^2 + (4-m) x + (3+2m) = 0$, with discriminant $\sqrt{(4-m)^2 - 4(3+2m)}$.

Suppose Case 1 has one solution and Case 2 has two. Then $m^2 + 8m + 16 - 12 + 8m = 0 \to m^2 + 16m + 4 = 0$ so $m = -8 \pm 2 \sqrt{15}$. The root of the quadratic is $\frac{-(m+4)}{2}$, and so $m = -8 + 2 \sqrt{15}$ works for the domain. This also gives valid roots in the other quadratic, so this is our solution.

Remark. If we instead suppose that Case 2 has the one solution we end up finding two values of $m$ which give invalid solutions for the Case 1 quadratic.

Answer 5

If you recognize that $x^2 + 4x + 3 = (x+2)^2 - 1,$ then it can quickly be seen that the graph of that function is an upward-opening parabola with its minimum at $x=-2$, where the value of the function is $-1$. Clearly for values of $x$ far from $-2$, the function is positive, those parts of the graph of the function are preserved when we take the absolute value, $|x^2 + 4x + 3|,$ but the part near $x=-2$ (bounded by the points where the parabola crosses the $x$ axis to the left and right of $x=-2$) is flipped over the $x$ axis; that is, the section of that parabola below the $x$ axis is replaced by the section of the parabola $1-(x+2)^2$ above the $x$ axis.

So the graph of $|x^2 + 4x + 3|$ starts at large positive $y$ and large negative $x$, goes down to the $x$ axis, then goes up over a "hill" and down again until it arrives at the $x$ axis again, and finally goes up indefinitely.

The solutions of $|x^2+4x+3|-mx+2m=0$ are the points where $$|x^2 + 4x + 3| = m(x - 2),$$ so we can look for a line that intercepts the $x$ axis at $x=2$ with slope $m$ and that intersects the graph of $|x^2 + 4x + 3|$ in exactly three points. It is clear that a positive $m$ will not work, and for zero or negative $m$ it is easy to get either two intersection points or four; but the only way to get three intersection points is for the line to be just tangent to the "hill" in the middle of the graph of $|x^2 + 4x + 3|$. (A slightly shallower slope gives four intersections; steeper gives two.)

So you need there to be a unique solution to $m(x-2) = 1-(x+2)^2$. Rather than do much more algebra on that form of the equation, we can observe that $1-(x+2)^2$ describes the part of the graph of $|x^2 + 4x + 3|$ where the $|\cdot|$ operation changed the sign of the formula, so we really want there to be one (double root) solution of $$-(x^2+4x+3)-mx+2m = -x^2 -(4+m)x - (3-2m)=0.$$ Using the $b^2-4ac$ formula for the discriminant, the discriminant is $(m+4)^2 - 4(3-2m)$ or simplified, $m^2 + 16m + 4$. This has two possible solutions: the larger (steeper) value of $m$ corresponds to a line that is tangent to the parabola $-(x^2+4x+3)$ somewhere to the right of $x=2$ and far below the $x$-axis, in other words, nowhere near the graph of $|x^2 + 4x + 3|$; the smaller (shallower) slope $m$ is the solution you want.

Answer 6

From the figure it is clear that the only possibility is when the red straight line of slope m and passing through the point (2, 0) is tangent to |f(x)|in a point of the interval [-3, -1] (otherwise there are none or two or four solutions). QUESTION: Take m=ax^2+b; this gives 〖Ax〗^3+Bx^2+Cx+D=0 with three solutions x_i=g_i (m) so x_1+x_2+x_3=-B/A and others. What about this?