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# Find the x-intercepts of x^4-7x^2-3.

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

Guest Feb 20, 2015

#2
+19206
+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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#1
+85644
+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
+19206
+10
$$\boxed{x^4-7x^2-3=0}\qquad \small{\text{ we set  z= x^2  with \boxed{x=\sqrt{z}} }}\\\\ \small{\text{we have  z^2-7z-3= 0 }}\\\\ \small{\text{ z_{1,2}=\dfrac{7\pm\sqrt{49-4\cdot (-3)} } {2} }}\\\\ \small{\text{ z_{1,2}=\dfrac{7\pm\sqrt{ 49+12 } } {2} = \dfrac{7\pm\sqrt{ 61 } } {2} }}\\\\ \small{\text{ z_1 = \dfrac{7+\sqrt{ 61 } } {2} \qquad z_2 = \dfrac{7-\sqrt{ 61 } } {2} }}\\\\ \small{\text{x = \pm \sqrt{z_{1,2}}  }}\\\\ \small{\text{x_1 = + \sqrt{z_{1}} = \sqrt{ \dfrac{7+\sqrt{ 61 } } {2} }  }}\\\\ \small{\text{x_2 = - \sqrt{z_{1}} = -\sqrt{ \dfrac{7+\sqrt{ 61 } } {2} }  }}\\\\ \small{\text{x_3 = + \sqrt{z_{2}} = \sqrt{ \dfrac{7-\sqrt{ 61 } } {2} } \qquad 7 - \sqrt{61} < 0 \quad ! \quad (non-real)  }}\\\\ \small{\text{x_4 = - \sqrt{z_{2}} = -\sqrt{ \dfrac{7-\sqrt{ 61 } } {2} } \qquad 7 - \sqrt{61} < 0 \quad ! \quad (non-real)  }}\\\\$$