How do you solve #x^210x+25=35#?
3 Answers
Answer:
or
Explanation:

Take 35 from both sides so that the quadratic equation is equal to 0.
#x^2 10x  10 = 0# 
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)#
Answer:
#x=5+sqrt35#
Explanation:
#color(blue)(x^210x+25=35#
This is a quadratic equation (in form of
We use the quadratic formula to solve it
#color(brown)(x=(b+sqrt(b^24ac))/(2a)#
Where,
#color(violet)(a=1#
#color(violet)(b=10#
#color(violet)(c=10#
Answer:
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:
#(x5)^2 = x^210x+25 = 35#
Take the square root of both ends, allowing for both positive and negative square roots to get:
#x5 = +sqrt(35)#
Add
#x = 5+sqrt(35)#
Finally note that