Let there be a quadratic expression given by \(y = a{x^2} + bx + c\) where \(a,b,c \in R\). We can transform this expression into a form where we may be able to draw certain conclusions about the behavior of the quadratic expression easily. Let us look at the form we are talking about. You may already be acquainted with completing the square method of finding the roots of the quadratic equation.
Take 'a' common from the expression:
\(a{x^2} + bx + c\)
\( = a\left( {{x^2} + \frac{b}{a}x + \frac{c}{a}} \right)\)
Now multiply and divide the second term of the expression by 2:
\( = a\left( {{x^2} + 2 \cdot \left( {\frac{b}{{2a}}} \right)x + \frac{c}{a}} \right)\)
Now add and subtract \({\frac{{{b^2}}}{{4{a^2}}}}\) . The purpose of which is to get a perfect square of the form \({\left( {x + y} \right)^2} = {x^2} + 2xy + {y^2}\).
We get,
\(a{x^2} + bx + c\)\( = \)\(a\left( {{x^2} + 2 \cdot \left( {\frac{b}{{2a}}} \right)x + \frac{{{b^2}}}{{4{a^2}}} - \frac{{{b^2}}}{{4{a^2}}} + \frac{c}{a}} \right)\)
Completing the square we get
\( = \)\(a\left[ {{{\left( {x + \frac{b}{{2a}}} \right)}^2} - \frac{{{b^2}}}{{4{a^2}}} + \frac{c}{a}} \right]\)
\( = \)\(a\left[ {{{\left( {x + \frac{b}{{2a}}} \right)}^2} - \frac{{{b^2} - 4ac}}{{4{a^2}}}} \right]\)
\( = \)\(a\left[ {{{\left( {x + \frac{b}{{2a}}} \right)}^2} - \frac{\Delta }{{4{a^2}}}} \right]\)
where \(\Delta = {b^2} - 4ac\) is called the discriminant of the quadratic expression.
This form of the expression now has a perfect square (the squared term) plus a constant. This form we will use to calculate range, roots, extremum and graph of the expression.
You should solve these examples to get a clear idea of the above process.
Example 1:
\({x^2} + x + 1\)
Solution:
\( = \)\({x^2} + x + 1\)
\( = \)\({x^2} + 2 \cdot \left( {\frac{1}{2}} \right)x + 1\)
\( = \)\({x^2} + 2 \cdot \left( {\frac{1}{2}} \right)x + \frac{1}{4} - \frac{1}{4} + 1\)
\( = \)\({\left( {x + \frac{1}{2}} \right)^2} + \frac{3}{4}\)
Example 2:
\(2{x^2} - 5x + 4\)
Solution:
\( = \)\(2\left( {{x^2} - \frac{5}{2}x + 2} \right)\)
\( = \)\(2\left( {{x^2} - 2 \cdot \left( {\frac{5}{4}} \right)x + 2} \right)\)
\( = \)\\(2\left( {{x^2} - 2 \cdot \left( {\frac{5}{4}} \right)x + \frac{{25}}{{16}} - \frac{{25}}{{16}} + 2} \right)\)
\( = \)\(2\left[ {{{\left( {x - \frac{5}{4}} \right)}^2} + \frac{7}{{16}}} \right]\)
Example 3:
\( - 5{x^2} + 9x - 6\)
Solution:
\(\begin{array}{l} = - 5\left( {{x^2} - \frac{9}{5}x + \frac{6}{5}} \right)\\ = - 5\left( {{x^2} - 2.\left( {\frac{9}{{10}}} \right)x + \frac{6}{5}} \right)\\ = - 5\left( {{x^2} - 2.\left( {\frac{9}{{10}}} \right)x + \frac{{81}}{{100}} - \frac{{81}}{{100}} + \frac{6}{5}} \right)\\ = - 5\left[ {{{\left( {x - \frac{9}{{10}}} \right)}^2} + \frac{{39}}{{100}}} \right]\end{array}\)
Take 'a' common from the expression:
\(a{x^2} + bx + c\)
\( = a\left( {{x^2} + \frac{b}{a}x + \frac{c}{a}} \right)\)
Now multiply and divide the second term of the expression by 2:
\( = a\left( {{x^2} + 2 \cdot \left( {\frac{b}{{2a}}} \right)x + \frac{c}{a}} \right)\)
Now add and subtract \({\frac{{{b^2}}}{{4{a^2}}}}\) . The purpose of which is to get a perfect square of the form \({\left( {x + y} \right)^2} = {x^2} + 2xy + {y^2}\).
We get,
\(a{x^2} + bx + c\)\( = \)\(a\left( {{x^2} + 2 \cdot \left( {\frac{b}{{2a}}} \right)x + \frac{{{b^2}}}{{4{a^2}}} - \frac{{{b^2}}}{{4{a^2}}} + \frac{c}{a}} \right)\)
Completing the square we get
\( = \)\(a\left[ {{{\left( {x + \frac{b}{{2a}}} \right)}^2} - \frac{{{b^2}}}{{4{a^2}}} + \frac{c}{a}} \right]\)
\( = \)\(a\left[ {{{\left( {x + \frac{b}{{2a}}} \right)}^2} - \frac{{{b^2} - 4ac}}{{4{a^2}}}} \right]\)
\( = \)\(a\left[ {{{\left( {x + \frac{b}{{2a}}} \right)}^2} - \frac{\Delta }{{4{a^2}}}} \right]\)
where \(\Delta = {b^2} - 4ac\) is called the discriminant of the quadratic expression.
This form of the expression now has a perfect square (the squared term) plus a constant. This form we will use to calculate range, roots, extremum and graph of the expression.
You should solve these examples to get a clear idea of the above process.
Example 1:
\({x^2} + x + 1\)
Solution:
\( = \)\({x^2} + x + 1\)
\( = \)\({x^2} + 2 \cdot \left( {\frac{1}{2}} \right)x + 1\)
\( = \)\({x^2} + 2 \cdot \left( {\frac{1}{2}} \right)x + \frac{1}{4} - \frac{1}{4} + 1\)
\( = \)\({\left( {x + \frac{1}{2}} \right)^2} + \frac{3}{4}\)
Example 2:
\(2{x^2} - 5x + 4\)
Solution:
\( = \)\(2\left( {{x^2} - \frac{5}{2}x + 2} \right)\)
\( = \)\(2\left( {{x^2} - 2 \cdot \left( {\frac{5}{4}} \right)x + 2} \right)\)
\( = \)\\(2\left( {{x^2} - 2 \cdot \left( {\frac{5}{4}} \right)x + \frac{{25}}{{16}} - \frac{{25}}{{16}} + 2} \right)\)
\( = \)\(2\left[ {{{\left( {x - \frac{5}{4}} \right)}^2} + \frac{7}{{16}}} \right]\)
Example 3:
\( - 5{x^2} + 9x - 6\)
Solution:
\(\begin{array}{l} = - 5\left( {{x^2} - \frac{9}{5}x + \frac{6}{5}} \right)\\ = - 5\left( {{x^2} - 2.\left( {\frac{9}{{10}}} \right)x + \frac{6}{5}} \right)\\ = - 5\left( {{x^2} - 2.\left( {\frac{9}{{10}}} \right)x + \frac{{81}}{{100}} - \frac{{81}}{{100}} + \frac{6}{5}} \right)\\ = - 5\left[ {{{\left( {x - \frac{9}{{10}}} \right)}^2} + \frac{{39}}{{100}}} \right]\end{array}\)
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