How do you solve #x^2-8x+5=0#?

2 Answers
Mar 15, 2018

Answer:

We use a process called completing the square, which works for all quadratic equations. Here it gives #x=4\pm \sqrt{11}#.

Explanation:

Rather than memorizing a formula, you should understand how "completing the square" works.

Render the equation as:

#x^2-8x=-5#

Then add a constant, which we will call #b^2#:

#color(blue)(x^2-8x+b^2=b^2-5)#

Now compare the left side with the identity

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

We then have

#x^2-8x+b^2=(a+b)^2#

if

#a^2=x^2# so set #a=x#

#2ab=-8x# so #b=-4#

So,go back to the equation in blue and put in #b=-4#, then:

#x^2-8x+16=(a+b)^2=(x-4)^2=11#

So #x-4=\pm \sqrt{11}# meaning

#x=4\pm \sqrt{11}#.

Mar 15, 2018

Answer:

Root 1 #color(blue)( = 4+sqrt(11)#

Root 2 #color(blue)( = 4-sqrt(11)#

Explanation:

Given:

#color(red)(x^2-8x+5=0#

The standard form of a quadratic is #color(blue)(ax^2+bx+c=0#

So, we have #color(blue)(a=1, b=-8 and c=5.#

Solutions are given by #color(red)([-b+-sqrt(b^2-4ac)]/(2a)#

#rArr [-(-8)+-sqrt((-8)^2-4(1)(5)]]/(2*1)#

Using scientific calculator, we get,

#rArr [8+-sqrt(44)]/(2# [ Intermediate Result 1 ]

Observe that #sqrt(44)# can also be simplified as

#sqrt(4*11)#

#rArr sqrt(2^2*11)#

#rArr sqrt(2^2)*sqrt(11)#

#rArr 2sqrt(11)#

Now, we can write [ Intermediate Result 1 ] as

#color(blue)((8+-2sqrt(11))/2]#

We can rewrite the above as

#color(blue)[8/2+-(2sqrt(11))/2]#

#(4*2)/2+-(2sqrt(11))/2#

#(4*cancel(2))/cancel(2)+-(cancel(2)sqrt(11))/cancel(2)#

#rArr 4+-sqrt(11)#

Hence, we have the following two solutions:

Root 1 #color(blue)( = 4+sqrt(11)#

Root 2 #color(blue)( = 4-sqrt(11)#

Using a scientific calculator, we can simplify the above results.

Root 1 #color(green)(~~ 7.31662479#

Root 2#color(green)(~~ 0.68337521#

You can verify the solutions visually by examining the image of the graph below:

enter image source here

Hope it helps.