How do you find the solution to the quadratic equation #x^2 - 4x -3 = 0#?

3 Answers
Apr 14, 2018

Answer:

#x=2+-sqrt7#

Explanation:

#"there are no whole numbers which multiply to - 3"#
#"and sum to - 4"#

#"we can solve using the method of "color(blue)"completing the square"#

#"the coefficient of the "x^2" term is 1"#

#• " add subtract "(1/2"coefficient of the x-term")^2" to"#
#x^2-4x#

#rArrx^2+2(-2)xcolor(red)(+4)color(red)(-4)-3=0#

#rArr(x-2)^2-7=0#

#rArr(x-2)^2=7#

#color(blue)"take the square root of both sides"#

#rArrx-2=+-sqrt7larrcolor(blue)"note plus or minus"#

#rArrx=2+-sqrt7larrcolor(red)"exact solutions"#

Apr 14, 2018

Answer:

x = #2+- sqrt(7)#

Explanation:

Apply quadratic formula for this equation instead of trying to factor it out.
1/ #((-b+-sqrt((b)^2-4(a)(c)))/(2(a)))#

2/ #((-(-4)+-sqrt((-4)^2-4(1)(-3)))/(2(1)))#

3/ #((4+-sqrt(16+12))/(2))#

4/ #((4+-2sqrt(7))/(2))# ( 2 cancel out)

5/ x = # 2+-sqrt(7)#

Apr 14, 2018

Answer:

#x=2+sqrt7 or x=2-sqrt7#

Explanation:

Here,

#x^2-4x-3=0#

#=>x^2-4x+4-7=0#

#=>(x-2)^2=7=(sqrt7)^2#

#=>x-2=+-sqrt7#

#=>x=2+-sqrt7#

OR

Comparing with quadratic equation,

#ax^2+bx+c=0=>a=1,b=-4,c=-3#

#triangle=b^2-4ac=(-4)^2-4(1)(-3)#

#=>triangle=16+12=28=4xx7#

#sqrt(triangle)=2sqrt7#

So,

#x=(-b+-sqrt(triangle))/(2a)#

#x=(4+-2sqrt7)/(2(1))#

#x=2+-sqrt7#