How do you solve #x^2-10x+25=35#?

3 Answers
Oct 2, 2016

Answer:

#x = 5 + sqrt(35)#
or
#x = 5 - sqrt(35)#

Explanation:

  1. Take 35 from both sides so that the quadratic equation is equal to 0.
    #x^2 -10x - 10 = 0#

  2. Solve for x using the following equation:
    #x = (-b +- sqrt(b^2 -4ac))/(2a)#
    #= (-(-10) +- sqrt((-10)^2 -4(1)(-10)))/(2(1))#
    #= (10 +- sqrt(100 +40))/2#
    #= (10 +- sqrt(140))/2#
    #= (10 +- (sqrt(35)*sqrt(4)))/2#
    #= (2(5) +- 2(sqrt(35)))/2#
    #= (2(5 +- sqrt(35)))/2#
    #= 5 +- sqrt(35)#

Oct 2, 2016

Answer:

#x=5+-sqrt35#

Explanation:

#color(blue)(x^2-10x+25=35#

#rarrx^2-10x+25-35=0#

#rarrx^2-10x-10=0#

This is a quadratic equation (in form of #color(orange)(ax^2+bx+c=0#)

We use the quadratic formula to solve it

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

Where,

#color(violet)(a=1#

#color(violet)(b=-10#

#color(violet)(c=-10#

#rarrx=(-(-10)+-sqrt((-10)^2-4(1)(-10)))/(2(1))#

#rarrx=(10+-sqrt(100-(-40)))/(2)#

#rarrx=(10+-sqrt(100+40))/(2)#

#rarrx=(10+-sqrt(140))/(2)#

#rarrx=(10+-sqrt(4*35))/(2)#

#rarrx=(10+-2sqrt(35))/(2)#

#rarrx=(cancel10^5+-cancel2^1sqrt(35))/(cancel2^1)#

#rArrcolor(green)(x=5+-sqrt35#

Oct 2, 2016

Answer:

#x = 5+-sqrt(35)#

Explanation:

Notice that the left hand side of the equation is already a perfect square trinomial.

So rather than reformulate, we can proceed as follows:

#(x-5)^2 = x^2-10x+25 = 35#

Take the square root of both ends, allowing for both positive and negative square roots to get:

#x-5 = +-sqrt(35)#

Add #5# to both sides to find:

#x = 5+-sqrt(35)#

Finally note that #35=5*7# has no square factors, so #sqrt(35)# is already in simplest form.