What is #5x^2-28x+19=0#?

2 Answers
Feb 21, 2018

See explanation below!

Explanation:

Recall that a linear equation in one variable is of the form #ax+b=0#, where #a# and #b# are constants and #a≠0#.

For example: #" "# #3x+5=0#

A quadratic equation has an #x^2# (x-squared) term. ("Quadratum" is Latin for square.) The general quadratic equation in standard form looks like:

#ax^2+bx+c=0# #\ \ \ # #\cdots##\ \ ##\cdots# where #a\ne 0#

If we want to find the #x# or #x's# that work, we might guess and substitute and hope we get lucky, or we might try one of these four methods:

We can solve graphically by equating the polynomial to #y# instead of to #0#, we get an equation whose graph is a parabola. The #x-\text{intercepts}# of the parabola (if any) correspond to the solutions of the original quadratic equation.

Feb 21, 2018

The solutions are #x=(14+-sqrt101)/5#.

Explanation:

One way to find the solutions to a quadratic is to use the quadratic formula:

#x=(-b+-sqrt(b^2-4ac))/(2a)#

Here's our quadratic:

#5x^2-28x+19=0#

The values are #a=5#, #b=-28#, and #c=19#. Plug in the values to the equation:

#x=(-b+-sqrt(b^2-4ac))/(2a)#

#color(white)x=(-(-28)+-sqrt((-28)^2-4(5)(19)))/(2(5))#

#color(white)x=(28+-sqrt((-28)^2-4(5)(19)))/10#

#color(white)x=(28+-sqrt(784-4(5)(19)))/10#

#color(white)x=(28+-sqrt(784-380))/10#

#color(white)x=(28+-sqrt(404))/10#

#color(white)x=(28+-sqrt(4*101))/10#

#color(white)x=(28+-sqrt(2^2*101))/10#

#color(white)x=(28+-2sqrt(101))/10#

#color(white)x=(14+-sqrt(101))/5#

This is as simplified as the answer gets. The two final solutions are:

#x=(14+sqrt101)/5#
and
#x=(14-sqrt101)/5#

Here's the graph of the function (with an altered scale):

graph{5x^2-28x+19[-3,8,-30,20]}