Question

Use the quadratic formula to solve x^2 -3x - 5 = 0. Round to two decimal places.

Answer 1

Answer:

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=\frac{3+\sqrt{29}}{2},\:x=\frac{3-\sqrt{29}}{2}$

Step-by-step explanation:

Given the equation

$x^2\:-3x\:-\:5\:=\:0$

solving with the quadratic formula

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=1,\:b=-3,\:c=-5$

$x_{1,\:2}=\frac{-\left(-3\right)\pm \sqrt{\left(-3\right)^2-4\cdot \:1\cdot \left(-5\right)}}{2\cdot \:1}$

$x_{1,\:2}=\frac{-\left(-3\right)\pm \sqrt{29}}{2\cdot \:1}$

separating the solutions

$x_1=\frac{-\left(-3\right)+\sqrt{29}}{2\cdot \:1},\:x_2=\frac{-\left(-3\right)-\sqrt{29}}{2\cdot \:1}$

solving

$x=\frac{-\left(-3\right)+\sqrt{29}}{2\cdot \:\:1}$

  $=\frac{3+\sqrt{29}}{2\cdot \:1}$

  $=\frac{3+\sqrt{29}}{2}$

also solving

$\:x=\frac{-\left(-3\right)-\sqrt{29}}{2\cdot \:\:1}$

  $=\frac{3-\sqrt{29}}{2\cdot \:1}$

  $=\frac{3-\sqrt{29}}{2}$

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=\frac{3+\sqrt{29}}{2},\:x=\frac{3-\sqrt{29}}{2}$