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# Challenge!

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Find the value(s) of m in the quadratic x2+x+m+1=0 such that there is only one solution.

Oct 26, 2017

#1
+10441
+3

Find the value(s) of m in the quadratic x2+x+m+1=0 such that there is only one solution.

Oct 26, 2017
#2
+22905
+3

Find the value(s) of m in the quadratic x2+x+m+1=0
such that there is only one solution.

$$\begin{array}{|lrcll|} \hline & x^2+1\cdot x+ (m+1) &=& 0 \\ & \text{ or } (x-x_1) (x-x_2) &=& 0 \quad & | \quad \text{one solution so } x_1 = x_2 \\ & (x-x_1) (x-x_1) &=& 0 \\ & (x-x_1)^2 &=& 0 \\ & x^2 \underbrace{-2x_1}_{=1}x +\underbrace{ x_1^2}_{=m+1} &=& 0 \quad & | \quad \text{compare the coefficients} \\\\ (1) & -2x_1 &=& 1 \\ & x_1 &=& -\frac{1}{2} \\\\ (2) & x_1^2 &=& m+1 \\ & \left(-\frac{1}{2}\right)^2 &=& m+1 \\ & \frac{1}{4} &=& m+1 \\ & \frac{1}{4}-1 &=& m \\ & \frac{1}{4}-\frac{4}{4} &=& m \\\\ &\mathbf{ -\frac{3}{4}} &\mathbf{=}& \mathbf{m} \\ \hline \end{array}$$

Oct 26, 2017
#3
+2343
+2

$$m=-\frac{3}{4}$$ is one solution. Omi67 and heureka both assumed that m must be a constant, but nothing specifies this restriction. For example, could be a linear term or a quadratic term.

The discriminant of a quadratic equation is the part of the quadratic formula underneath the square root symbol (which is known as the radicand). By definition, the discriminant provides information about the roots of a given equation.

$$x = {-b \pm \sqrt{\textcolor{red}{b^2-4ac}}\over 2a}$$

For quadratics, there are 3 possible cases, and the different results convey something different.

1) If $$b^2-4ac<0$$, then there are no real roots.

2) If $$b^2-4ac=0$$, then there is one and only one real root.

3) If $$b^2-4ac>0$$, then there are two real and distinct roots.

This information will be utilized for the following problem. The second case is relevant here.

First, assume that is a constant. Before continuing, one should identify the values of a,b, and c

$$a=1\hspace{5mm}b=1\hspace{5mm}c=m+1$$

 $$b^2-4ac=0$$ Substitute in the known values for the variables. $$1^2-4(1)(m+1)=0$$ Simplify the left hand side. $$1-4(m+1)=0$$ Subtract 1 from both sides. $$-4(m+1)=-1$$ Divide by -4 on both sides. $$m+1=\frac{1}{4}$$ Subtract 1 from both sides to isolate m. $$m=-\frac{3}{4}$$

Of course, the other users also got the same answer. Now, one must consider the 2nd case:

$$a=1\hspace{5mm}b=m+1\hspace{5mm}c=1$$

$$(m+1)^2-4*1*1=0$$ Simplify the left hand side by doing the multiplication.
$$(\textcolor{green}{m+1})^2-\textcolor{blue}{4}=0$$ Notice that the left hand side is a difference of two squares; factor accordingly.
$$(\textcolor{green}{m+1}+\textcolor{blue}{2})(\textcolor{green}{m+1}-\textcolor{blue}{2})=0$$ Simplify both the factors.
$$(m+3)(m-1)=0$$ Set both factors equal to 0 and then solve each factor separately.
 $$m+3=0$$ $$m-1=0$$

Add the additive inverse of the constant in both equations to both sides.
 $$m=-3$$ $$m=1$$

One must be careful, though, because one solved for the b, the coefficient of the x-term (linear term). This means that $$m=-3x\hspace{1mm}\text{and}\hspace{1mm}m=x$$.

Time to consider the 3rd and final case.

$$a=m+1\hspace{1cm}b=1\hspace{1cm}c=1$$

 $$1^2-4(m+1)(1)=0$$ Wait a second! This looks exactly the same as our first case. One already knows, without solving, that m=-3/4

In this case m = coefficient of the x^2-term (quadratic term), so $$m=\frac{1}{4}x^2$$

Therefore, all the solutions are $$m=-\frac{3}{4}x^2,-3x,x,-\frac{3}{4}$$

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Oct 26, 2017