How do you solve #x^2-10x+9=0#?

2 Answers
Dec 13, 2016

Answer:

#x = 9# and #x = 1#

Explanation:

To factor you must play with factors which multiply to #9#
#(1xx9, " "3xx3, " " 9xx1)#

#(x - 9)(x - 1) = 0#

Now, solve each factor for #0#:

#x - 9 = 0#

#x - 9 + 9 = 0 +9#

#x - 0 = 9#

#x = 9#

and

#x - 1 = 0#

#x - 1 + 1 = 0 + 1#

#x - 0 = 1#

#x = 1#

Dec 13, 2016

Answer:

#x=1" "# or #" "x=9#

Explanation:

Given:

#x^2-10x+9 = 0#

#color(white)()#
Sum of coefficients shortcut

Notice that the sum of the coefficients is #0#.

That is:

#1-10+9 = 0#

Hence #x=1# is a solution and #(x-1)# a factor:

#x^2-10x+9 = (x-1)(x-9)#

There are several ways to spot that the other factor must be #(x-9)#. For example, the coefficient of #x# must be #1# so that when multiplied by the #x# in #(x-1)# results in #x^2# and the constant term must be #-9# so that when multiplied by #-1# gives #+9#.

So the other solution is #x=9#

#color(white)()#
Completing the square

Another method, which is a little over the top for this particular problem, involves completing the square, then using the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

with #a=(x-5)# and #b=4# as follows:

#0 = x^2-10x+9#

#color(white)(0) = x^2-10x+25-16#

#color(white)(0) = (x-5)^2-4^2#

#color(white)(0) = ((x-5)-4)((x-5)+4)#

#color(white)(0) = (x-9)(x-1)#

Hence solutions #x=9# and #x=1#

#color(white)()#
Quadratic formula

For completeness, I should also mention the quadratic formula.

The equation:

#x^2-10x+9=0#

is in the form

#ax^2+bx+c=0#

with #a=1#, #b=-10# and #c=9#

This has solutions given by the quadratic formula:

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

#color(white)(x) = (10+-sqrt((-10)^2-4(1)(9)))/(2*1)#

#color(white)(x) = (10+-sqrt(100-36))/2#

#color(white)(x) = (10+-sqrt(64))/2#

#color(white)(x) = (10+-8)/2#

#color(white)(x) = 5+-4#

That is:

#x = 5+4 = 9" "# or #" "x = 5-4 = 1#