How do you solve # x^2-9x+8=0#?

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Answer 1 · 2016-03-12T06:33:48

#x=8,1#

#color(blue)(x^2-9x+8=0#

This is a Quadratic equation (in form #ax^2+bx+c=0#)

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

Where

#color(blue)(a=1,b=-9,c=8#

Substitute the values in the formula

#rarrx=(-(-9)+-sqrt(-9^2-4(1)(8)))/(2(1))#

#rarrx=(9+-sqrt(81-32))/(2)#

#rarrx=(9+-sqrt(49))/(2)#

#rarrx=(9+-7)/(2)#

Now, we have #2# values for #x#

#1)color(orange)(x=(9+7)/2#

#2)color(indigo)(x=(9-7)/2#

Solve

#1)color(green)(x=(9+7)/2=16/2=8#

#2)color(green)(x=(9-7)/2#