Quadratic Function Formula
A quadratic expression can always be factorized, but the factorization process may be difficult if the zeroes of the expression are non-integer real numbers, or non-real numbers. In such cases, we can use the quadratic formula to determine the zeroes of the expression. The geeral form of a quadratic function is given as:
f(x) = ax2 + bx + c,
where a, b, and c are numbers with a not equal to zero and constants.
What is Quadratic Function Formula?
Consider an arbitrary quadratic equation:
\[a{x^2} + bx + c = 0,\,\,\,a \ne 0\]
To determine the roots of this equation, we proceed as follows:
\[\begin{align}&a{x^2} + bx = - c\\&\Rightarrow \,\,\,{x^2} + \frac{b}{a}x = - \frac{c}{a}\end{align}\]
Now, we express the left hand side as a perfect square, by introducing a new term on both sides:
\[{x^2} + \frac{b}{a}x + \underbrace {{{\left( {\frac{b}{{2a}}} \right)}^2}}_{{\rm{New}}\,{\rm{term}}} = - \frac{c}{a} + \underbrace {{{\left( {\frac{b}{{2a}}} \right)}^2}}_{{\rm{New term}}}\]
The left hand side is now a perfect square:
\[{\left( {x + \frac{b}{{2a}}} \right)^2} = - \frac{c}{a} + \frac{{{b^2}}}{{4{a^2}}} = \frac{{{b^2} - 4ac}}{{4{a^2}}}\]
This is good for us, because now we can take square roots to obtain:
\[\begin{align}&x + \frac{b}{{2a}} = \pm \frac{{\sqrt {{b^2} - 4ac} }}{{2a}}\\&\Rightarrow \,\,\,x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\end{align}\]
Thus, by completing the squares, we were able to isolate \(x\) and obtain the two roots of the equation. Let us apply this formula to a few examples.
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Example 1:
Using the quadratic formula, find the roots of the equation \({x^2} - 7x + 6 = 0\).
Solution:
Comparing the given equation to the standard quadratic form \(a{x^2} + bx + c = 0\), we have:
\[a = 1,\,\,b =\;\; - 7,\,\,c = 6\]
Now, we use the quadratic formula to find the roots:
\[\begin{align}&x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\&\Rightarrow \,\,\,x = \frac{{ - \left( { - 7} \right) \pm \sqrt {{{\left( { - 7} \right)}^2} - 4\left( 1 \right)\left( 6 \right)} }}{{2\left( 1 \right)}}\\& = \frac{{7 \pm \sqrt {49 - 24} }}{2} = \frac{{7 \pm \sqrt {25} }}{2}\\& = \frac{{7 \pm 5}}{2} = 1,\,\,6\end{align}\]
Thus, the two roots are \(x = 1\) and \(x = 6\). This can be verified by factorization as well. The given equation can be factorized as follows:
\[\begin{align}&{x^2} - 7x + 6 = 0\\&\Rightarrow \,\,\,{x^2} - 6x - x + 6 = 0\\&\Rightarrow \,\,\,x\left( {x - 6} \right) - 1\left( {x - 6} \right) = 0\\&\Rightarrow \,\,\,\left( {x - 1} \right)\left( {x - 6} \right) = 0\end{align}\]
Answer: The two roots are \(x = 1\) and \(x = 6\).
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Example 2:
Find the roots of the following equation in the variable x:
\[2p\left( {1 + {x^2}} \right) - \left( {1 + {p^2}} \right)\left( {x + p} \right) = 0\]
Solution:
Writing the given equation in the standard quadratic form, we have (verify this):
\[2p{x^2} - \left( {1 + {p^2}} \right)x + p - {p^3} = 0\]
The coefficients are:
\[a = 2p,\,\,b = - \left( {1 + {p^2}} \right),\,\,c = p - {p^3}\]
Applying the quadratic formula, we have:
\[\begin{align}&\Rightarrow \,\,\,x = \frac{{\left( {1 + {p^2}} \right) \pm \sqrt {{{\left( {1 + {p^2}} \right)}^2} - 4\left( {2p} \right)\left( {p - {p^3}} \right)} }}{{2\left( {2p} \right)}}\\& = \frac{{\left( {1 + {p^2}} \right) \pm \sqrt {{p^4} + 2{p^2} + 1 - 8p\left( {p - {p^3}} \right)} }}{{4p}}\\&= \frac{{\left( {1 + {p^2}} \right) \pm \sqrt {9{p^4} - 6{p^2} + 1} }}{{4p}}\\& = \frac{{\left( {1 + {p^2}} \right) \pm \sqrt {{{\left( {3{p^2} - 1} \right)}^2}} }}{{4p}}\end{align}\]
Now, we take the square root to obtain the two roots:
\[\begin{align}&x = \frac{{\left( {1 + {p^2}} \right) \pm \left( {3{p^2} - 1} \right)}}{{4p}}\\& = \frac{{4{p^2}}}{{4p}},\,\,\,\,\frac{{2 - 2{p^2}}}{{4p}}\\& = p,\,\,\,\frac{{1 - {p^2}}}{{2p}}\end{align}\]
Answer: The roots are \(p,\,\,\,\dfrac{{1 - {p^2}}}{{2p}}\)
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