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Root equations
An equation with a linear term in the unknown #x# and one term as #\sqrt{ax^2+bx+c}# for certain numbers #a# and #b# can be solved by
- First, isolating the term #\sqrt{ax^2+bx+c}# to the left hand side of the equal sign; the form of the equation then is #\sqrt{ax^2+bx+c} = px+q#;
- next, taking the square of the left and right hand side; the form of the equation then is #ax^2+bx+c=(px+q)^2#;
- next, solve the quadratic equation;
- Finally, determine whether #a\cdot e^2 + b\cdot e + c\ge0# and #p\cdot e + q\ge0# for each solution #x=e# of the quadratic equation; If so, then #x=e# is a solution to the original equation; if not, then we need to drop #x=e#.
The quadratic equation has not more than two solutions. These are not necessarily solutions to the initial equations. A solution #x=e# of the quadratic equation is, after all, only a solution if #a\cdot e^2 + b\cdot e + c\ge0# and #p\cdot e + q\ge0#.
In the last step we still have to check for inequalities, because the equation resulting from taking the square in the second step, is no longer equivalent with the original equation. After all, it is possible that #x=e# satisfies the equation #ax^2+bx+c=(px+q)^2# and #p\cdot e+q\lt 0#, but then #x=e# is not a solution to the original equation. For example, #x=3# is not a solution to \[ \sqrt{x^2+16}=2x-11\] but it is to \[ x^2+16=(2x-11)^2\]
However, any solution to the equation is a also a solution to the equation that arises squaring both members.
To see this, we reduce the equation to a quadratic equation in the following manner.
\[ \begin{array}{rclcl}
\sqrt{17-x^2}&=& -(x+5) &\phantom{xxxx}&\color{blue}{\text{isolate the root}}\\
17-x^2 &=& (x+5)^2 &\phantom{xxxx}&\color{blue}{\text{taking the square}}\\
17-x^2 &=& x^2+10\cdot x+25 &\phantom{xxxx}&\color{blue}{\text{removing the brackets }}\\
-2\cdot x^2-10\cdot x-8 &=& 0 &\phantom{xxxx}&\color{blue}{\text{all terms to the left hand side }}\\
x^2+5\cdot x+4 &=& 0 &\phantom{xxxx}&\color{blue}{\text{dividing by }-2\tiny.}\\
\end{array}\]
The solution to this quadratic equation is #x=-4\lor x=-1#.
To make sure this is a solution to the original equation #x=-4# must satisfy #17-x^2\ge0# and #-(x+5)\ge0#. This is ikke the case. The same conditions hold for #x=-1#. These are ikke fulfilled.
Hence, as solution to the original equation, we find: #ingen#.
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