How do you solve #x^2+9=10x#?

2 Answers
Apr 11, 2016

1 and 9

Explanation:

#y = x^2 - 10x + 9 = 0.#
Find 2 real roots, knowing sum (-b = 10) and product (c = 9).
They are 1 and 9.

Apr 11, 2016

#9,1#

Explanation:

#color(blue)(x^2+9=10x#

Subtract #10x# both sides

#rarrx^2+9-10x=cancel(10x-10x#

#rarrx^2+9-10x=0#

Rewrite the equation in Standard form (#ax^2+bx+c=0#)

#color(purple)(rarrx^2-10x+9=0#

This is a Quadratic formula

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

Remember that #aand b# are the coefficients and #c# is the constant

So,

#color(violet)(a=1#

#color(violet)(b=-10#

#color(violet)(c=9#

#rarrx=(-(-10)+-sqrt(-10^2-4(1)(9)))/(2(1))#

#rarrx=(10+-sqrt(100-(36)))/(2)#

#rarrx=(10+-sqrt(64))/(2)#

#rarrx=(10+-8)/(2)#

#rarrx=(cancel10^5+-cancel8^4)/(cancel2^1)#

#rarrx=5+-4#

Remember that #+-# means "plus or minus",

Which implies that,

#color(indigo)(x=5+4=9#

#color(orange)(x=5-4=1#

#color(blue)( ul bar |x=9,1|#

If you are confused with the Quadratic formula

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