What is #5x^2-28x+19=0#?
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#2|k^2# then #2|k# for some #k\inZZ#"
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.
The solutions are #x=(14+-sqrt101)/5#.
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]}
