How do you solve #x^2-10x+16=0# using the quadratic formula?

2 Answers
Mar 20, 2018

Answer:

#x=8 or x=2#

Explanation:

We know the Quadratic Formula:

If #color(red)(ax^2+bx+c=0) ,#

then, #color(red)(x=(-b+-sqrt(triangle))/(2a)and triangle=b^2-4ac)#

Comparing , #color(blue)(x^2-10x+16=0#, with quad.equn ,we get

#a=1 ,b=-10, c=16#

First we find #triangle#

#triangle=(-10)^2-4(1)(16)=100-64=36#

#=>sqrt(triangle)=sqrt(36)=6#

So,

#x=(-b+-sqrt(triangle))/(2a)=(-(-10)+-6)/(2(1))=(10+-6)/2#

#i.e.x=(10+6)/2 or x=(10-6)/2#

#=>x=16/2 or x=4/2#

#=>x=8 or x=2#

Mar 20, 2018

Answer:

#x = 8,2#

Explanation:

Your equation is already in quadratic form, so all we have to do is substitute. The quadratic form is as follows:
#ax^2 + bx + c = 0#

Your equation looks like this:
#x^2 - 10x + 16 = 0#

What we are trying to do is find the values of #a, b# and #c#. They are located in the same spot in your equation as they are in the quadratic form.

#ax^2 + bx + c = 0#

#(1)x^2 + (-10)x + (16) = 0#
#a = 1#
#b = -10#
#c = 16#

Now plug these in to the quadratic formula:
# x = (-b \pm sqrt(b^2-4ac)) / (2a) #

# x = (-(-10) \pm sqrt(-10^2-4(1)(16)))/ (2(1)) #

# x = ((color(red)+10) \pm sqrt(color(red)100-(color(blue)64)))/ (2) #

# x = (10 \pm sqrt(100-64))/ (2) #

# x = (10 \pm sqrt(36))/ (2) #

# x = (10 \pm 6)/ (2) #

Now solve by doing addition alone and subtraction alone to get your two values:

# x = (10 color(red)+ 6)/ (2) rarr x = (16)/ (2) rarr x = 8 #

# x = (10 color(red)- 6)/ (2) rarr x = 4/2 rarr x = 2 #

So the solution is #x = 8,2#