Question

How do you find the roots of ${{x}^{3}}+{{x}^{2}}-17x+15=0$

Answer 1

Hint: First, try to substitute values like 1, -1 and 0 as they are the most common roots in this kind of questions. As it turns out, 1 satisfies the given equation and is a root so (x-1) will be a factor. Now, we add and subtract terms to remove the factor (x-1) from ${{x}^{3}}+{{x}^{2}}-17x+15=0$ . After taking the factor common we will have the equation $(x-1)({{x}^{2}}+2x-15)=0$ . Now, we say, that \[x=1\] and ${{x}^{2}}+2x-15=0$ are the solutions or roots of the equation. Simplify the quadratic equation further by either using the quadratic formulae, that is \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a};\text{ for the equation }a{{x}^{2}}+bx+c=0\] or by the method of completing the square. In the following solution, we will use the method of completing the square.

Complete step by step answer:
We have been asked to find the roots of ${{x}^{3}}+{{x}^{2}}-17x+15=0$
First, we will substitute 1 in ${{x}^{3}}+{{x}^{2}}-17x+15$ to check whether that satisfies the equation:
$\Rightarrow {{1}^{3}}+{{1}^{2}}-17\left( 1 \right)+15=0$
Thus, 1 satisfies the equation.
Hence, 1 is a root of the given equation.
$\Rightarrow \text{ }$ (x-1) is a factor of ${{x}^{3}}+{{x}^{2}}-17x+15$ .
Now, we add and subtract terms to factor ${{x}^{3}}+{{x}^{2}}-17x+15=0$
$\Rightarrow {{x}^{3}}\underline{-{{x}^{2}}+{{x}^{2}}}+{{x}^{2}}-17x+15=0$
\[\Rightarrow {{x}^{3}}-{{x}^{2}}+2{{x}^{2}}-2x-15x+15=0\Rightarrow {{x}^{2}}(x-1)+2x(x-1)-15(x-1)=0\]
Now, taking (x-1) common, we have:
\[\Rightarrow (x-1)({{x}^{2}}+2x-15)=0\]
From the above equation we have that x will then take the values for which \[x-1=0\] and ${{x}^{2}}+2x-15=0$ .
That is x=1 and the solutions for the equation ${{x}^{2}}+2x-15=0$
Now, we are going to use the completing the square method.
That is, we will add and subtract 4 to left-hand side of ${{x}^{2}}+2x-15=0$
$\Rightarrow {{x}^{2}}+2x+4-19=0\Rightarrow {{\left( x+2 \right)}^{2}}=19$
Taking root on both sides we have:
$\Rightarrow x+2=\pm \sqrt{19}$
$\Rightarrow x=-2\pm \sqrt{19}$

Therefore, the 3 roots to the given equation ${{x}^{3}}+{{x}^{2}}-17x+15=0$ are $1,-2+\sqrt{19},-2-\sqrt{19}$ .

Note: The above method is called completing the square because we add and subtract terms to make it in the form of a whole square equal to a constant. Alternatively, you can solve ${{x}^{2}}+2x-15=0$ directly using the quadratic formula, $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . If using the above method, be careful while taking the root on both sides after completing the square as $\pm $ is missed often.