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How do I find the x-intercepts of x^4-7x^2-3?

Guest Feb 20, 2015

Best Answer 

 #2
avatar+18829 
+10

How do I find the x-intercepts of  ?

$$\boxed{x^4-7x^2-3=0}\qquad
\small{\text{
we set $ z= x^2 $ with $\boxed{x=\sqrt{z}}
$}}\\\\
\small{\text{we have $ z^2-7z-3= 0 $}}\\\\
\small{\text{$
z_{1,2}=\dfrac{7\pm\sqrt{49-4\cdot (-3)} } {2}
$}}\\\\
\small{\text{$
z_{1,2}=\dfrac{7\pm\sqrt{ 49+12 } } {2} = \dfrac{7\pm\sqrt{ 61 } } {2}
$}}\\\\
\small{\text{$
z_1 = \dfrac{7+\sqrt{ 61 } } {2} \qquad z_2 = \dfrac{7-\sqrt{ 61 } } {2}
$}}\\\\
\small{\text{$x = \pm \sqrt{z_{1,2}} $ }}\\\\
\small{\text{$x_1 = + \sqrt{z_{1}} = \sqrt{ \dfrac{7+\sqrt{ 61 } } {2} } $ }}\\\\
\small{\text{$x_2 = - \sqrt{z_{1}} = -\sqrt{ \dfrac{7+\sqrt{ 61 } } {2} } $ }}\\\\
\small{\text{$x_3 = + \sqrt{z_{2}} = \sqrt{ \dfrac{7-\sqrt{ 61 } } {2} } \qquad 7 - \sqrt{61} < 0 \quad ! \quad (non-real)
$ }}\\\\
\small{\text{$x_4 = - \sqrt{z_{2}} = -\sqrt{ \dfrac{7-\sqrt{ 61 } } {2} } \qquad 7 - \sqrt{61} < 0 \quad ! \quad (non-real)
$ }}\\\\$$

heureka  Feb 20, 2015
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2+0 Answers

 #1
avatar+81032 
+5

The y-intercept is (0, -3)

To find the x-intercepts, set the function to 0

x^4-7x^2-3 = 0      this will not factor....let's use the on-site solver

$${{\mathtt{x}}}^{{\mathtt{4}}}{\mathtt{\,-\,}}{\mathtt{7}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{3}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{{\sqrt{{\sqrt{{\mathtt{61}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{7}}}}}{{\sqrt{{\mathtt{2}}}}}}\\
{\mathtt{x}} = {\frac{{\sqrt{{\sqrt{{\mathtt{61}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{7}}}}}{{\sqrt{{\mathtt{2}}}}}}\\
{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{{\sqrt{{\sqrt{{\mathtt{61}}}}{\mathtt{\,-\,}}{\mathtt{7}}}}{\mathtt{\,\times\,}}{i}}{{\sqrt{{\mathtt{2}}}}}}\\
{\mathtt{x}} = {\frac{{\sqrt{{\sqrt{{\mathtt{61}}}}{\mathtt{\,-\,}}{\mathtt{7}}}}{\mathtt{\,\times\,}}{i}}{{\sqrt{{\mathtt{2}}}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{2.721\: \!235\: \!902\: \!665\: \!060\: \!4}}\\
{\mathtt{x}} = {\mathtt{2.721\: \!235\: \!902\: \!665\: \!060\: \!4}}\\
{\mathtt{x}} = -{\mathtt{0.636\: \!494\: \!177\: \!470\: \!090\: \!6}}{i}\\
{\mathtt{x}} = {\mathtt{0.636\: \!494\: \!177\: \!470\: \!090\: \!6}}{i}\\
\end{array} \right\}$$

So....we have two "real" x intercepts and two "non-real" solutions......note....because of the symmetry of the solutions......... this graph is also symmetric to the origin   

This graph confirms this......https://www.desmos.com/calculator/xtuatwy6b7

 

CPhill  Feb 20, 2015
 #2
avatar+18829 
+10
Best Answer

How do I find the x-intercepts of  ?

$$\boxed{x^4-7x^2-3=0}\qquad
\small{\text{
we set $ z= x^2 $ with $\boxed{x=\sqrt{z}}
$}}\\\\
\small{\text{we have $ z^2-7z-3= 0 $}}\\\\
\small{\text{$
z_{1,2}=\dfrac{7\pm\sqrt{49-4\cdot (-3)} } {2}
$}}\\\\
\small{\text{$
z_{1,2}=\dfrac{7\pm\sqrt{ 49+12 } } {2} = \dfrac{7\pm\sqrt{ 61 } } {2}
$}}\\\\
\small{\text{$
z_1 = \dfrac{7+\sqrt{ 61 } } {2} \qquad z_2 = \dfrac{7-\sqrt{ 61 } } {2}
$}}\\\\
\small{\text{$x = \pm \sqrt{z_{1,2}} $ }}\\\\
\small{\text{$x_1 = + \sqrt{z_{1}} = \sqrt{ \dfrac{7+\sqrt{ 61 } } {2} } $ }}\\\\
\small{\text{$x_2 = - \sqrt{z_{1}} = -\sqrt{ \dfrac{7+\sqrt{ 61 } } {2} } $ }}\\\\
\small{\text{$x_3 = + \sqrt{z_{2}} = \sqrt{ \dfrac{7-\sqrt{ 61 } } {2} } \qquad 7 - \sqrt{61} < 0 \quad ! \quad (non-real)
$ }}\\\\
\small{\text{$x_4 = - \sqrt{z_{2}} = -\sqrt{ \dfrac{7-\sqrt{ 61 } } {2} } \qquad 7 - \sqrt{61} < 0 \quad ! \quad (non-real)
$ }}\\\\$$

heureka  Feb 20, 2015

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