Quadratic Equations: Variations
Substitution
In linear equation with one unknown, we saw that certain equations can be reduced to linear equations by substitution of the unknown. This also applies to quadratic equations. Below some examples are given.
#x=\frac{1}{2}\sqrt{2} \lor x=2 \lor x=-\frac{1}{2}\sqrt{2} \lor x=-2#
With the substitution #y=x^{2}# we find #2y^2-9 y+4=0#. First, we solve this equation in #y# with the abc-formula: \[ \begin{array}{rclcl}2y^2-9 y+4&=&0&\phantom{x}&\color{blue}{\text{after substitution } y=x^{2}}\\y&=&\frac{+9\pm\sqrt{49}}{2\cdot 2}&\phantom{x}&\color{blue}{\text{abc-formula}}\\ y={{1}\over{2}} &\lor & y=4 &\phantom{x}&\color{blue}{\text{simplified}}\end {array}\]
The solutions for #x# follow from the relation #x=\sqrt{y}\lor x=-\sqrt{y}#, where we must keep in mind that #\sqrt{y}# is only defined for #y\ge0# (See Higher Power Roots). The solution for #x# is \[x=\sqrt{{{1}\over{2}}} \lor x=\sqrt{4} \lor x=-\sqrt{{{1}\over{2}}} \lor x=-\sqrt{4}\tiny.\] Hence, the answer is in standard form for roots, \[x=\frac{1}{2}\sqrt{2} \lor x=2 \lor x=-\frac{1}{2}\sqrt{2} \lor x=-2\tiny.\]
With the substitution #y=x^{2}# we find #2y^2-9 y+4=0#. First, we solve this equation in #y# with the abc-formula: \[ \begin{array}{rclcl}2y^2-9 y+4&=&0&\phantom{x}&\color{blue}{\text{after substitution } y=x^{2}}\\y&=&\frac{+9\pm\sqrt{49}}{2\cdot 2}&\phantom{x}&\color{blue}{\text{abc-formula}}\\ y={{1}\over{2}} &\lor & y=4 &\phantom{x}&\color{blue}{\text{simplified}}\end {array}\]
The solutions for #x# follow from the relation #x=\sqrt{y}\lor x=-\sqrt{y}#, where we must keep in mind that #\sqrt{y}# is only defined for #y\ge0# (See Higher Power Roots). The solution for #x# is \[x=\sqrt{{{1}\over{2}}} \lor x=\sqrt{4} \lor x=-\sqrt{{{1}\over{2}}} \lor x=-\sqrt{4}\tiny.\] Hence, the answer is in standard form for roots, \[x=\frac{1}{2}\sqrt{2} \lor x=2 \lor x=-\frac{1}{2}\sqrt{2} \lor x=-2\tiny.\]
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