How do you use the quadratic formula to solve for x-intercepts $x^{2} - 5 x - 3 = 0$ $x^2 − 5x − 3 = 0$ ?

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Answer 1 · 2017-06-13T04:24:57

$x \approx 5.54138 or \approx - 0.54138$ $x approx 5.54138 or approx -0.54138$

The quadratic formula states that for a quadratic equation of standard form $(a x^{2} + b x + c = 0)$ $(ax^2+bx+c=0)$ it's roots are given by:

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b+-sqrt(b^2-4ac))/(2a)$

In this example we are asked to find the x-intercepts (which are the roots) of $x^{2} - 5 x - 3 = 0$ $x^2-5x-3 =0$

$:. a=1, b=-5, c=-3$

Hence $x= (-(-5)+-sqrt((-5)^2-4*1*(-3)))/(2*1)$

$= (5+-sqrt(25+12))/2$

$= (5+-sqrt(37))/2$

$x approx 5.54138 or approx -0.54138$

These x-intercepts can be seen on the graph of this quadratic below

graph{x^2-5x-3 [-18.14, 22.4, -9.91, 10.36]}

How do you use the quadratic formula to solve for x-intercepts x^2 − 5x − 3 = 0x2−5x−3=0x^2 − 5x − 3 = 0?