7.2 Quadratic Equation
\(ax^2+bx+c=0\) is a quadratic equation, then its roots are given by \(x=\frac {-b+\sqrt {b^2-4ac}}{2a}\) or \(x=\frac {-b-\sqrt {b^2-4ac}}{2a}\)
Solution.
In this case it is difficult to factorise \(2x^2-5x+1\), we then use the formula to solve the equation \(2x^2-5x+1=0\).
We have that \(a=2, b=-5, c=1\). Thus \begin {align*} x= &\frac {-(-5)+\sqrt {(-5)^2-4(2)(1)}}{2(2)}=\frac {5+\sqrt {17}}{4}\\ &\text {or}\\ x = & \frac {-(-5)-\sqrt {(-5)^2-4(2)(1)}}{2(2)}=\frac {5-\sqrt {17}}{4}\\ \end {align*}
Now, let \(\alpha \) and \(\beta \) be the roots of the equation \(ax^2+bx+c=0\). Then by the formula \(\alpha = \frac {-b+\sqrt {b^2-4ac}}{2a}\) and \(\beta =\frac {-b-\sqrt {b^2-4ac}}{2a}\).
Adding the two roots gives \begin {align*} \alpha + \beta & = \frac {-b+\sqrt {b^2-4ac}}{2a}+\frac {-b-\sqrt {b^2-4ac}}{2a}\\\\ & = \frac {-b+\sqrt {b^2-4ac}-b-\sqrt {b^2-4ac}}{2a}\\ & = \frac {-2b}{2a}\\\\ \therefore \quad \alpha + \beta & =\frac {-b}{a}\\ \end {align*}
Multiplying the two roots we obtain \begin {align*} \alpha \beta & = \left (\frac {-b+\sqrt {b^2-4ac}}{2a}\right )\left (\frac {-b-\sqrt {b^2-4ac}}{2a}\right )\\\\ & = \frac {b^2+b\sqrt {b^2-4ac}-b\sqrt {b^2-4ac}-(b^2-4ac)}{4a^2}\\\\ & = \frac {4ac}{4a^2}=\frac {c}{a}\\\\ \therefore \quad \alpha \beta & = \frac {c}{a}\\ \end {align*}
Thus, if \(\alpha \) and \(\beta \) are the roots of the equation \(ax^2+bx+c=0\), then \(\alpha +\beta =\frac {-b}{a}\) and \(\alpha \beta =\frac {c}{a}\).
Note 7.3. These two identities are more useful than they look, because they give the sum and the product without solving the equation. Any symmetric expression in the roots — one unchanged when \(\alpha \) and \(\beta \) are swapped — can be rebuilt from them, and the examples that follow do exactly that: \[\alpha ^{2}+\beta ^{2}=(\alpha +\beta )^{2}-2\alpha \beta ,\quad \frac {1}{\alpha }+\frac {1}{\beta }=\frac {\alpha +\beta }{\alpha \beta },\quad (\alpha -\beta )^{2}=(\alpha +\beta )^{2}-4\alpha \beta .\]
Two things follow. First, this works even when the roots are complex: the sum and product stay real, so a question like the one below on \(2x^{2}-x+3=0\) can be answered in ordinary arithmetic although neither root is a real number.
Second, the identities run backwards. Given any two numbers \(p\) and \(q\), the equation with those roots is \[x^{2}-(p+q)x+pq=0,\] which is how a new quadratic is built from old roots without ever finding them. Note \((\alpha -\beta )^{2}=(\alpha +\beta )^{2}-4\alpha \beta \) is \(\frac {D}{a^{2}}\) — the discriminant again, measuring how far apart the roots lie.
Let \(\alpha \) and \(\beta \) be the roots of the equation \(3x^2-7x+1=0\).
Find
- 1.
- \(\alpha +\beta \)
- 2.
- \(\alpha \beta \)
Solution.
- 1.
- \(\alpha +\beta = \frac {-b}{a}=\frac {-(-7)}{3}=\frac {7}{3}\)
- 2.
- \(\alpha \beta = \frac {c}{a}=\frac {1}{3}\)
\begin {align*} (\alpha + \beta )^2 &= (\alpha +\beta )(\alpha +\beta )\\ &= \alpha ^2+\alpha \beta + \alpha \beta +\beta ^2\\ & = \alpha ^2+2\alpha \beta + \beta ^2\\ &= \alpha ^2+\beta ^2+2\alpha \beta \\\\ \implies \quad \alpha ^2+\beta ^2 & = (\alpha +\beta )^2-2\alpha \beta \\ \end {align*}
Let \(\alpha \) and \(\beta \) be the roots of the equation \(7x^2+2x-5=0\). Find
-
- (a)
- \(\frac {1}{\alpha }+\frac {1}{\beta }\)
- (b)
- \(\alpha ^2+\beta ^2\)
- (c)
- \(\frac {\alpha }{\beta }+\frac {\beta }{\alpha }\)
- (d)
- \(\alpha -\beta \)
Solution.
\(7x^2+2x-5=0\)
\(\therefore \quad a=7, b=2, c=-5\)
\(\implies \quad \alpha +\beta = \frac {-b}{a}=\frac {-2}{7}\)
\(\implies \quad \alpha \beta =\frac {c}{a}=\frac {-5}{7}\)
-
- (a)
- \(\frac {1}{\alpha }+\frac {1}{\beta }=\frac {\beta +\alpha }{\alpha \beta }=\frac {-\frac {2}{7}}{-\frac {5}{7}}=\frac {2}{5}\)
- (b)
- \(\alpha ^2+\beta ^2\) using the identity \(\alpha ^2+\beta ^2=(\alpha +\beta )^2-2\alpha \beta \), we have that
\(\alpha ^2+\beta ^2=\left (\frac {-2}{7}\right )^2-2\left (\frac {-5}{7}\right )=\frac {4}{49}+\frac {10}{7}=\frac {4+70}{49}\)
\(\therefore \quad \alpha ^2+\beta ^2=\frac {74}{49}\)
- (c)
- \(\frac {\alpha }{\beta }+\frac {\beta }{\alpha }=\frac {\alpha ^2+\beta ^2}{\alpha \beta }=\frac {74/49}{-5/7}=-\frac {74}{35}\)
- (d)
- \(\alpha -\beta \) \begin {align*} \text {Note that}\quad (\alpha -\beta )^2 & = \alpha ^2-2\alpha \beta +\beta ^2\\ & = (\alpha ^2+\beta ^2)-2\alpha \beta \\ & = (\alpha + \beta )^2-2\alpha \beta -2\alpha \beta \\ & = (\alpha + \beta )^2-4\alpha \beta \\ \end {align*}
Therefore, \((\alpha -\beta )^2=\left (\frac {-2}{7}\right )^2-4\left (\frac {-5}{7}\right )=\frac {4}{49}+\frac {20}{7}=\frac {4+140}{49}=\frac {144}{49}\) \[\therefore \quad \alpha -\beta =\pm \sqrt {\frac {144}{49}}\] \(\implies \quad \alpha -\beta =\frac {12}{7}\) or \(\alpha -\beta =-\frac {12}{7}\).
Let \(\alpha \) and \(\beta \) be the roots of the equation \(2x^2-x+3=0\). Find an equation whose roots are \(\alpha ^2+1\) and \(\beta ^2+1\).
\[2x^2-x+3=0\]
\[a=2, b=-1, c=3\]
\[\alpha +\beta =-\frac {b}{a}=\frac {1}{2}\quad \text {and}\quad \alpha \beta = \frac {c}{a}=\frac {3}{2}\]
The roots of the new equation are \(p=\alpha ^2+1\) and \(q=\beta ^2+1\). We need to find their sum \(p+q\) and their product \(pq\).
The new equation is of the form \(x^2-(p+q)x+pq=0\). Now, \begin {align*} p+q & =(\alpha ^2+1)+(\beta ^2+1)\\ & = \alpha ^2+\beta ^2+2\\ & = (\alpha +\beta )^2-2\alpha \beta +2\\ & = \left (\frac {1}{2}\right )^2-2\left (\frac {3}{2}\right )+2\\ & = \frac {1}{4}-3+2=\frac {1}{4}-1\\ & = \frac {-3}{4} \end {align*}
\begin {align*} pq & =(\alpha ^2+1)(\beta ^2+1)\\ & = \alpha ^2(\beta ^2+1)+1(\beta ^2+1)\\ & = (\alpha \beta )^2+\alpha ^2+\beta ^2+1\\ & = (\alpha \beta )^2+(\alpha +\beta )^2-2\alpha \beta +1\\\\ \therefore \quad pq & = \left (\frac {3}{2}\right )^2+\left (\frac {1}{2}\right )^2-2\left (\frac {3}{2}\right )+1\\ & = \frac {9}{4}+\frac {1}{4}-3+1\\ & = \frac {10-8}{4}\\ & = \frac {1}{2}\\ \end {align*}
Thus the equation is \begin {align*} x^2-\left (\frac {-3}{4}\right )x+\frac {1}{2} & = 0\\ x^2+\frac {3}{4}x+\frac {1}{2} & = 0\\ \implies \quad 4x^2+3x+2 & =0\\ \end {align*}
Given the equation \(3x^2+8x+d=0\). Find the value of \(d\) if the roots of this equation differ by 2.
Solution.
Let \(\alpha \) and \(\beta \) be the roots of \(3x^2+8x+d=0\) then \(\alpha +\beta =\frac {-8}{3}\) and \(\alpha \beta =\frac {d}{3}\).
Now, \(\alpha - \beta = 2\)
- \(\alpha +\beta =\frac {-8}{3}\cdots (1)\)
- \(\alpha -\beta = 2\cdots (2)\)
\begin {align*} \implies \quad \beta & = \alpha -2\\\\ \implies \quad \alpha +\alpha -2 & = \frac {-8}{3}\\\\ \implies \quad 2\alpha = \frac {-8}{3}+2 & =\frac {-8+6}{3}\\ \end {align*}
\[\therefore \quad \alpha =\frac {-1}{3}\]
\begin {align*} \frac {-1}{3}-\beta &=2\\\\ \implies \quad \beta =\frac {-1}{3}-2 & = \frac {-1-6}{3}\\\\ \therefore \quad \beta &=\frac {-7}{3}\\ \end {align*}
\begin {align*} \alpha \beta & = \frac {d}{3}\\\\ \left (\frac {-1}{3}\right )\left (\frac {-7}{3}\right ) & = \frac {d}{3}\\\\ \frac {7}{9} & = \frac {d}{3} \implies \quad d=\frac {7}{3}\\\\ \end {align*}
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