Section1.11The quadratic formula and the discriminant
ΒΆSubsection1.11.1Written Examples
The solutions to a quadratic equation of form
\begin{equation*} ax^2+bx+c=0 \end{equation*}can be found using the formula
\begin{equation*} x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\text{.} \end{equation*}The solution equation is called the quadratic formula.
For example, let's use the quadratic formula to solve the equation
\begin{equation*} x^2-4x-32=0\text{.} \end{equation*}We begin by identifying the constants \(a\text{,}\) \(b\text{,}\) and \(c\text{.}\)
\begin{equation*} a=1, b=-4, c=-32\text{.} \end{equation*}We now replace the constants in the quadratic formula with their respective values and simplify the result.
\begin{align*} x\amp=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-(-4) \pm \sqrt{(-4)^2-4 \cdot 1 \cdot -32}}{2 \cdot 1}\\ x\amp=\frac{4 \pm \sqrt{144}}{2}\\ x\amp=\frac{4 \pm 12}{2} \end{align*}The symbol "\(\pm\)" is read "plus or minus" and implies that there are two solutions - one when the symbol is executed as subtraction and another when the symbol is executed as addition. Picking up where we left off.
\begin{align*} x\amp=\frac{4-12}{2}\amp\amp\text{or}\amp x\amp=\frac{4+12}{2}\\ x\amp=-4\amp\amp\text{or}\amp x\amp=8 \end{align*}So the solutions to the equation are \(-4\) and \(8\) and the solution set is \(\{-4, 8\}\text{.}\)
As just demonstrated, one can use the quadratic formula without having an understanding of its origin. The remainder of this section is devoted to the derivation of the quadratic formula. If you'd rather just get on with using it, go ahead to the next section.
The derivation of the quadratic equation relies on the process of completing the square. We being with the equation
\begin{equation*} ax^2+bx+c=0, a \neq 0\text{.} \end{equation*}Before completing the square we need to isolate the constant and divide both sides by \(a\) so that the coefficient on the squared term is \(1\text{.}\)
\begin{align*} ax^2+bx+c\amp=0\\ ax^2+bx\amp=-c\\ \frac{1}{a} \cdot (ax^2+bx)\amp=\frac{1}{a} \cdot -c\\ x^2+\frac{b}{a}x\amp=-\frac{c}{a} \end{align*}We now complete the square on the left side of the equation by adding the square of half the linear coefficient. This action is balanced by adding the same expression to the right side of the equation. We'll then go ahead and combine the terms on the right side of the equation.
\begin{align*} x^2+\frac{b}{a}x+(\frac{b}{2a})^2\amp=-\frac{c}{a}+(\frac{b}{2a})^2\\ x^2+\frac{b}{a}x+\frac{b^2}{4a^2}\amp=-\frac{c}{a}+\frac{b^2}{4a^2}\\ x^2+\frac{b}{a}x+\frac{b^2}{4a^2}\amp=-\frac{c}{a} \cdot \frac{4a}{4a}+\frac{b^2}{4a^2}\\ x^2+\frac{b}{a}x+\frac{b^2}{4a^2}\amp=-\frac{4ac}{4a^2}+\frac{b^2}{4a^2}\\ x^2+\frac{b}{a}x+\frac{b^2}{4a^2}\amp=\frac{-4ac+b^2}{4a^2}\\ x^2+\frac{b}{a}x+\frac{b^2}{4a^2}\amp=\frac{b^2-4ac}{4a^2} \end{align*}We now factor the left side into its perfect square and complete the derivation using the square root method. Note that when we simplify \(\sqrt{4a^2}\) we can simply state \(2a\text{.}\) It's true that we don't know whether \(a\) is a positive number or a negative number, but when we extract the roots we introduce the \(\pm\) sign, so we will have two roots, one positive and one negative, regardless of whether \(a\) is positive or negative (assuming that \(b^2-4ac \neq 0\)). If \(b^2-4ac=0\text{,}\) then \(\frac{b^2-4ac}{4a^2}=\frac{0}{2a}=0\) whether \(a\) is positive or negative.
\begin{align*} x^2+\frac{b}{a}x+\frac{b^2}{4a^2}\amp=\frac{b^2-4ac}{4a^2}\\ (x+\frac{b}{2a})^2\amp=\frac{b^2-4ac}{4a^2}\\ x+\frac{b}{2a}\amp=\pm \sqrt{\frac{b^2-4ac}{4a^2}}\\ x+\frac{b}{2a}\amp=\pm \frac{\sqrt{b^2-4ac}}{2a}\\ x\amp=-\frac{b}{2a}\pm \frac{\sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \end{align*}Let's use the quadratic formula to solve the equation \(2x^2-5x=7\text{.}\)
We begin by making the right side of the equation zero and stating the values of \(a\text{,}\) \(b\text{,}\) and \(c\text{.}\)
\begin{align*}
2x^2-5x\amp=7\\
2x^2-5x-7\amp=0
\end{align*}
\begin{gather*}
\\
a=2, b=-5, c=-7
\end{gather*}
Finally, we state our solutions and solution set.
The solutions are \(-1\) and \(\frac{7}{2}\text{.}\) The solution set is \(\{-1,\frac{7}{2}\}\)
When the square root expression does not simplify to an integer, there's no need to spit the solution into two separate equations, although we do need to make sure that we state the solutions as separate numbers (assuming that there are two solutions). We also need to make sure that we completely simplify the square root and the fraction.
Let's use the quadratic formula to solve the equation \(x^2+10x+5=0\text{.}\)
The equation is already in standard form, so we can go ahead and state the values of \(a\text{,}\) \(b\text{,}\) and \(c\) and proceed with the quadratic formula.
\begin{align*}
x^2+10x+5\amp=0
\end{align*}
\begin{gather*}
a=1, b=10, c=5
\end{gather*}
The solutions are \(-5-2\sqrt{5}\) and \(-5+2\sqrt{5}\text{.}\)
The solution set is \(\{-5-2\sqrt{5}, -5+2\sqrt{5}\}\text{.}\)
When the ultimate value under the square root symbol is zero, the equation has only one solution. This is because \(\sqrt{0}=0\text{,}\) and it doesn't alter the result changing subtraction of zero to addition of zero.
When the ultimate value under the square root symbol is negative, the equation has no real number solutions because square roots of negative numbers are not real numbers (although they really are numbers). In many situations we are only interested in real number solutions, so we stop the solution process once the negative radicand has been identified and state that there are no real number solutions.
Difficult as it may be to believe, there are real life applications where numbers with imaginary parts (square roots of negative numbers) are meaningful. Because of this, we do need to examine some equations where the solutions are complex numbers with imaginary parts. We do this in the next section.
Let's solve the quadratic equation \(4x^2-12x+25=0\text{.}\) The equation is already in standard form, so we can state the values of \(a\text{,}\) \(b\text{,}\) and \(c\) and proceed with the quadratic formula.
\(4x^2-12x+25=0\)
\(a=4, b=-12, c=25\)
If we were only concerned with real number solutions we could state our conclusion now, because \(\sqrt{-256}\) is not a real number. However, let's assume that we our also interested in solutions that include imaginary parts and proceed. In our stated solutions we want the real part of each number separate from the imaginary part, so we will simplify towards that end.
\begin{align*} x\amp=\frac{12 \pm \sqrt{-256}}{8}\\ x\amp=\frac{12 \pm 16i}{8}\\ x\amp=\frac{12}{8} \pm \frac{16}{8}i\\ x\amp=\frac{3}{2} \pm 2i \end{align*}The solutions are \(\frac{3}{2}-2i\) and \(\frac{3}{2}+2i\text{.}\)
The solution set is \(\{\frac{3}{2}-2i,\frac{3}{2}+2i\}\text{.}\)
Let's see another example.
Use the quadratic formula to solve \(x(2x-5)=4(x^2+3)\text{.}\)
We need to manipulate the equation into standard form before we state the values of \(a\text{,}\) \(b\text{,}\) and \(c\text{.}\) I prefer that \(a\) is a positive number, so after I expand both sides I am going to move all the terms to the right side of the equation.
\begin{gather*}
x(2x-5)=4(x^2+3)\\
2x^2-5x=4x^2+12\\
0=2x^2+5x+12
\end{gather*}
\begin{gather*}
\\
\\
a=2, b=5, c=12
\end{gather*}
The solutions are \(-\frac{5}{4}-\frac{\sqrt{71}}{4}i\) and \(-\frac{5}{4}+\frac{\sqrt{71}}{4}i\text{.}\)
The solution set is \(\{-\frac{5}{4}-\frac{\sqrt{71}}{4}i,-\frac{5}{4}+\frac{\sqrt{71}}{4}i\}\text{.}\)
There are times when we are concerned about the nature of the solutions to a quadratic equation rather than the precise solutions themselves. For example, when the equation \(ax^2+bx+c=0\) has two, one, or zero real number solutions, we know that the graph of \(y=ax^2+bx+c\) has, respectively, two, one, or zero \(x\)-intercepts.
Consider the quadratic formula.
\begin{equation*} x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \end{equation*}Because \(\sqrt{b^2-4ac}\) is the expression that is both added and subtracted, and it the part of the expression that can be an imaginary number, it is our focus when determining the nature of the solution set. We can narrow our focus even further, because the nature is determined by the value of \(b^2-4ac\text{.}\) When that value is positive, the equation will have two real number solutions. When the value is zero there will be one real number solution, and when the value is negative there will be two complex number solutions each with imaginary parts.
Because the expression \(b^2-4ac\) plays the key role in the nature of the solution set to a quadratic equation, it is given a name. The expression is called the determinant. Let's consider three examples.
Example
For the equation \(x^2-4x-7=0\text{,}\) \(a=1\text{,}\) \(b=-4\text{,}\) and \(c=-7\text{.}\) The discriminant is derived below.
\begin{align*} b^2-4ac\amp=(-4)^2-4 \cdot 1\cdot -7\\ \amp=44 \end{align*}Because \(b^2-4a \gt 0\text{,}\) we know that the equation \(x^2-4x-7=0\) has two real number solutions and that the graph of \(y=x^2-4-7\) has two \(x\)-intercepts.
Example
For the equation \(2x^2-6x+9=0\text{,}\) \(a=2\text{,}\) \(b=-6\text{,}\) and \(c=9\text{.}\) The discriminant is derived below.
\begin{align*} b^2-4ac\amp=(-6)^2-4 \cdot 2 \cdot 9\\ \amp=-36 \end{align*}Because \(b^2-4ac \lt 0\text{,}\) we know that the equation \(2x^2-6x+9=0\) has no real number solutions and that the graph of \(y=2x^2-6x+9\) has no \(x\)-intercepts. The equation \(2x^2-6x+9=0\) has two complex number solutions, each of which has a non-zero imaginary term.
Example
For the equation \(4x^2-12x+9=0\text{,}\) \(a=4\text{,}\) \(b=-12\text{,}\) and \(c=9\text{.}\) The discriminant is derived below.
\begin{align*} b^2-4ac\amp=(-12)^2-4 \cdot 4 \cdot 9\\ \amp=0 \end{align*}Because \(b^2-4ac=0\text{,}\) we know that the equation \(4x^2-12x+9=0\) has exactly one real number solution and that the graph of \(y=4x^2-12x+9\) has exactly one \(x\)-intercept.
Subsection1.11.2Practice Exercises
Use the quadratic formula to solve each of the following equations over the real numbers. For each equation state both the solutions and the solution set.
- \(x^2+6x-4=0\)
- \(t^2+10t=39\)
- \(3x^2-2x+5=0\)
- \(2x(3x-5)=5\)
- \(1-4x^2=x^2+10x+6\)
Solution
-
The quadratic equation is stated in standard form, so we can apply the quadratic formula immediately.
\begin{align*} x\amp=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-6 \pm \sqrt{6^2-4 \cdot 1 \cdot -4}}{2 \cdot 1}\\ x\amp=\frac{-6 \pm \sqrt{52}}{2}\\ x\amp=\frac{-6 \pm 2\sqrt{13}}{2}\\ x\amp=\frac{2(-3 \pm \sqrt{13})}{2}\\ x\amp=-3 \pm \sqrt{13} \end{align*}\(x^2+6x-4=0\)
\(a=1, b=6, c=-4\)
The solutions are \(-3-\sqrt{3}\) and \(-3+\sqrt{13}\text{.}\)
The solution set is \(\{-3-\sqrt{13}, -3+\sqrt{13}\}\text{.}\)
-
We need to make one side of the equation zero before we can apply the quadratic formula.
\begin{align*} x\amp=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-10 \pm \sqrt{10^2-4 \cdot 1 \cdot -39}}{2 \cdot 1}\\ x\amp=\frac{-10 \pm \sqrt{256}}{2}\\ x\amp=\frac{-10 \pm 16}{2} \end{align*} \begin{align*} x\amp=\frac{-10-16}{2}\amp\amp\text{or}\amp x\amp=\frac{-10+16}{2}\\ x\amp=-13\amp\amp\text{or}\amp x\amp=3 \end{align*}\begin{align*} t^2+10t\amp=39\\ t^2+10t-39\amp=0 \end{align*}\begin{gather*} \\ a=1, b=10, c=-10 \end{gather*}The solutions are \(-13\) and \(3\text{.}\)
The solution set is \(\{-13, 3\}\text{.}\)
-
The quadratic equation is stated in standard form, so we can apply the quadratic formula immediately.
\begin{align*} x\amp=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-(-2) \pm \sqrt{(-2)^2-4 \cdot 3 \cdot 5}}{2 \cdot 3}\\ x\amp=\frac{2 \pm \sqrt{-56}}{2} \end{align*}\(3x^2-2x+5=0\)
\(a=3, b=-2, c=5\)
Over the real numbers, there is no square root of \(-56\text{.}\) So over the real numbers the given equation has no solutions and the solution set is \(\emptyset\text{.}\)
-
We need to expand the left side of the equation and make one side of the equation zero before we can apply the quadratic formula.
\begin{align*} x\amp=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-(-10) \pm \sqrt{(-10)^2-4 \cdot 6 \cdot -5}}{2 \cdot 6}\\ x\amp=\frac{10 \pm \sqrt{220}}{12}\\ x\amp=\frac{10 \pm 2\sqrt{55}}{12}\\ x\amp=\frac{2(5 \pm \sqrt{55})}{2 \cdot 6}\\ x\amp=\frac{5 \pm \sqrt{55}}{6} \end{align*}\begin{align*} 2x(3x+5)\amp=5\\ 6x^2-10x\amp=5\\ 6x^2-10x-5\amp=0 \end{align*}\begin{gather*} \\ \\ a=6, b=-10, c=5 \end{gather*}The solutions are \(\frac{5-\sqrt{55}}{6}\) and \(\frac{5+\sqrt{55}}{6}\text{.}\)
The solution set is \(\{\frac{5-\sqrt{55}}{6}, \frac{5+\sqrt{55}}{6}\}\text{.}\)
-
We need to make one side of the equation zero before we can apply the quadratic formula. We choose to make the left side of the equation zero so that the squared term has a positive coefficient. Note that we can divide out a factor of five from both sides of the equation before applying the quadratic formula. Making this choice makes the numbers we're working with much more managable.
\begin{align*} x\amp=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-2 \pm \sqrt{2^2-4 \cdot 1 \cdot 1}}{2 \cdot 1}\\ x\amp=\frac{-2 \pm \sqrt{0}}{2}\\ x\amp=-1 \end{align*}\begin{align*} 1-4x^2\amp=x^2+10x+6\\ 0\amp=5x^2+10x+5\\ \frac{1}{5} \cdot 0\amp=\frac{1}{5} \cdot (5x^2+10x+5)\\ 0\amp=x^2+2x+1 \end{align*}\begin{gather*} \\ \\ \\ \\ a=1, b=2, c=1 \end{gather*}The only solution is \(1\) and the solution set is \(\{1\}\text{.}\)
Use the square root method to solve each of the following equations. Make sure that all solutions have been completely simplified. State both the solutions and the solution set.
- \((x+6)^2=-9\)
- \((3x-8)^2=-20\)
- \(x^2-12x+52=0\)
- \(3w^2-12w+24=0\)
Use the quadratic formula to solve each of the following equations. Make sure that all solutions have been completely simplified. State both the solutions and the solution set.
- \(4x^2+2x+1=0\)
- \(2t^2-3t+5=0\)
- \(3x^2+16=0\)
- \(y^2-4y+9=0\)
Solution
-
We can immediately apply the square root property.
\begin{align*} (x+6)^2\amp=-9\\ x+6\amp=\pm \sqrt{-9}\\ x+6\amp=\pm 3i\\ x\amp=-6\pm 3i \end{align*}The solutions are \(-6-3i\) and \(-6+3i\text{.}\)
The solution set is \(\{-6-3i, -6+3i\}\text{.}\)
-
We can immediately apply the square root property.
\begin{align*} (3x-8)^2\amp=-20\\ 3x-8\amp=\pm \sqrt{-20}\\ 3x-8\amp=\pm 2\sqrt{5}i\\ 3x\amp=8\pm 2\sqrt{5}i\\ x\amp=\frac{8\pm 2\sqrt{5}i}{3}\\ x\amp=\frac{8}{3} \pm \frac{2\sqrt{5}}{3}i \end{align*}The solutions are \(\frac{8}{3}-\frac{2\sqrt{5}}{3}i\) and \(\frac{8}{3}+\frac{2\sqrt{5}}{3}i\text{.}\)
The solution set is \(\frac{8}{3}-\frac{2\sqrt{5}}{3}i, \frac{8}{3}+\frac{2\sqrt{5}}{3}i\text{.}\)
-
We need to complete the square before we can employ the square root method.
\begin{align*} x^2-12x+52\amp=0\\ x^2-12x\amp=-52\\ x^2-12x+36\amp=-52+36\\ (x-6)^2\amp=-16\\ x-6\amp=\sqrt{-16}\\ x-6\amp=4i\\ x\amp=6\pm 4i \end{align*}The solutions are \(6-4i\) and \(6+4i\text{.}\)
The solution set is \(\{6-4i, 6+4i\}\text{.}\)
-
We need to simplify and complete the square before employing the square root method.
\begin{align*} 3w^2-12w+24\amp=0\\ \frac{1}{3} \cdot (3w^2-12w+24)\amp=\frac{1}{3} \cdot 0\\ w^2-4w+8\amp=0\\ w^2-4w\amp=-8\\ w^2-4w+4\amp=-8+4\\ (w-2)^2\amp=-4\\ w-2\amp=\pm \sqrt{-4}\\ w-2\amp=2i\\ w\amp=2\pm 2i \end{align*}The solutions are \(2-2i\) and \(2+2i\text{.}\)
The solution set is \(\{2-2i, 2+2i\}\text{.}\)
-
The equation is already in standard form, so we may proceed directly to the quadratic formula.
\begin{align*} x\amp=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\\ x\amp\frac{-2 \pm \sqrt{2^2-4 \cdot 4 \cdot 1}}{2 \cdot 4}\\ x\amp=\frac{-2 \pm \sqrt{-12}}{8}\\ x\amp=\frac{-2 \pm 2\sqrt{3}i}{8}\\ x\amp=-\frac{2}{8} \pm \frac{2\sqrt{3}}{8}i\\ x\amp=-\frac{1}{4} \pm \frac{\sqrt{3}}{4}i \end{align*}\(4x^2+2x+1=0\)
\(a=4, b=2, c=1\)
The solutions are \(-\frac{1}{4}-\frac{\sqrt{3}}{4}i\) and \(-\frac{1}{4}+\frac{\sqrt{3}}{4}i\text{.}\)
The solutions set is \(-\frac{1}{4}-\frac{\sqrt{3}}{4}i, -\frac{1}{4}+\frac{\sqrt{3}}{4}i\)
-
The equation is already in standard form, so we may proceed directly to the quadratic formula.
\begin{align*} x\amp=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-(-3) \pm \sqrt{(-3)^2-4 \cdot 1 \cdot 5}}{2 \cdot 1}\\ x\amp=\frac{3 \pm \sqrt{-11}}{2}\\ x\amp=\frac{3 \pm \sqrt{11}i}{2}\\ x\amp=\frac{3}{2} \pm \frac{\sqrt{11}}{2}i \end{align*}\(t^2-3t+5=0\)
\(a=1, b=-3, c=5\)
The solutions are \(\frac{3}{2}-\frac{\sqrt{11}}{2}i\) and \(\frac{3}{2}+\frac{\sqrt{11}}{2}i\text{.}\)
The solutions set is \(\{\frac{3}{2}-\frac{\sqrt{11}}{2}i, \frac{3}{2}+\frac{\sqrt{11}}{2}i\}\)
-
The equation is already in standard form, so we may proceed directly to the quadratic formula.
\begin{align*} x\amp=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-0 \pm \sqrt{0^2-4 \cdot 3 \cdot 16}}{2 \cdot 3}\\ x\amp=\frac{\pm \sqrt{-192}}{6}\\ x\amp=\pm \frac{8\sqrt{3}i}{6}\\ x\amp=\pm \frac{2 \cdot 4\sqrt{3}i}{2 \cdot 3}\\ x\amp=\pm \frac{4\sqrt{3}}{3}i \end{align*}\(3x^2+16=0\)
\(a=3, b=0, c=16\)
The solutions are \(-\frac{4\sqrt{3}}{3}i\) and \(\frac{4\sqrt{3}}{3}i\text{.}\)
The solutions set is \(\{-\frac{4\sqrt{3}i}{3}, \frac{4\sqrt{3}i}{3}\}\text{.}\)
-
The equation is already in standard form, so we may proceed directly to the quadratic formula.
\begin{align*} x\amp=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-(-4) \pm \sqrt{(-4)^2-4 \cdot 1 \cdot 9}}{2 \cdot 1}\\ x\amp=\frac{4 \pm \sqrt{-20}}{2}\\ x\amp=\frac{4 \pm 2\sqrt{5}i}{2}\\ x\amp=\frac{4}{2} \pm \frac{2\sqrt{5}}{2}i\\ x\amp=2 \pm \sqrt{5}i \end{align*}\(y^2-4y+9=0\)
\(a=1, b=-4, c=9\)
The solutions are \(2-\sqrt{5}i\) and \(2+\sqrt{5}i\text{.}\)
The solutions set is \(\{2-\sqrt{5}i, 2+\sqrt{5}i\}\)
For each equation, use the value of the discriminant to determine the nature of the solution. That's all - do not find the actual solutions.
- \(3x^2+4x-7=0\)
- \(36-12x-x^2=0\)
- \(x^2+4=0\)
- \(3x(3x-16)=-64\)
- \(7x^2=15x\)
Solution
-
\begin{align*} b^2-4ac\amp=4^2-4 \cdot 3 \cdot -7\\ \amp=100 \end{align*}
\(3x^2+4x-7=0\)
\(a=3, b=4, c=-7\)
Because the discriminant is a positive number, we know that the given equation has two real number solutions.
-
\begin{align*} b^2-4ac\amp=(-12)^2-4 \cdot -1 \cdot 36\\ \amp=0 \end{align*}
\(36-12x-x^2=0\)
\(a=-1, b=-12, c=36\)
Because the discriminant is zero, we know that the given equation has exactly one real number solution.
-
\begin{align*} b^2-4ac\amp=0^2-4 \cdot 1 \cdot 4\\ \amp=-16 \end{align*}
\(x^2+4=0\)
\(a=1, b=0, c=4\)
Because the discriminant is a negative number, we know that the given equation has two complex number solutions with non-zero imaginary parts
-
We need to write the equation in standard form before we can determine the determinant.
\begin{align*} b^2-4ac\amp=(-48)^2-4 \cdot 9 \cdot 64\\ \amp=0 \end{align*}\begin{align*} 3x(3x-16)\amp=-64\\ 9x^2-48x\amp=-64\\ 9x^2-48x+64\amp=0 \end{align*}\begin{gather*} \\ \\ a=9, b=-48, c=64 \end{gather*}Because the discriminant is zero, we know that the given equation has one real number solution.
-
We need to write the equation in standard form before we can determine the determinant.
\begin{align*} b^2-4ac\amp=(-15)^2-4 \cdot 7 \cdot 0\\ \amp=225 \end{align*}\begin{align*} 7x^2\amp=15x\\ 7x^2-15x\amp=0 \end{align*}\begin{gather*} \\ a=7, b=-15, c=0 \end{gather*}Because the discriminant is positive, the given equation has two real number solutions.