Find the value(s) of m in the quadratic x2+x+m+1=0 such that there is only one solution.
+12528 Find the value(s) of m in the quadratic x2+x+m+1=0 such that there is only one solution.
+26367 Find the value(s) of m in the quadratic x2+x+m+1=0
such that there is only one solution.
\(\begin{array}{|lrcll|} \hline & x^2+1\cdot x+ (m+1) &=& 0 \\ & \text{ or } (x-x_1) (x-x_2) &=& 0 \quad & | \quad \text{one solution so } x_1 = x_2 \\ & (x-x_1) (x-x_1) &=& 0 \\ & (x-x_1)^2 &=& 0 \\ & x^2 \underbrace{-2x_1}_{=1}x +\underbrace{ x_1^2}_{=m+1} &=& 0 \quad & | \quad \text{compare the coefficients} \\\\ (1) & -2x_1 &=& 1 \\ & x_1 &=& -\frac{1}{2} \\\\ (2) & x_1^2 &=& m+1 \\ & \left(-\frac{1}{2}\right)^2 &=& m+1 \\ & \frac{1}{4} &=& m+1 \\ & \frac{1}{4}-1 &=& m \\ & \frac{1}{4}-\frac{4}{4} &=& m \\\\ &\mathbf{ -\frac{3}{4}} &\mathbf{=}& \mathbf{m} \\ \hline \end{array}\)
+2441 \(m=-\frac{3}{4}\) is one solution. Omi67 and heureka both assumed that m must be a constant, but nothing specifies this restriction. For example, m 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 m 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. | ||
| Add the additive inverse of the constant in both equations to both sides. | ||
| |||
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 \(m=-\frac{3}{4}x^2,-3x,x,-\frac{3}{4}\)
