How do you find the solution to the quadratic equation x^2 - 4x -3 = 0x24x3=0?

3 Answers
Apr 14, 2018

x=2+-sqrt7x=2±7

Explanation:

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

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

"the coefficient of the "x^2" term is 1"the coefficient of the x2 term is 1

• " add subtract "(1/2"coefficient of the x-term")^2" to" add subtract (12coefficient of the x-term)2 to
x^2-4xx24x

rArrx^2+2(-2)xcolor(red)(+4)color(red)(-4)-3=0x2+2(2)x+443=0

rArr(x-2)^2-7=0(x2)27=0

rArr(x-2)^2=7(x2)2=7

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

rArrx-2=+-sqrt7larrcolor(blue)"note plus or minus"x2=±7note plus or minus

rArrx=2+-sqrt7larrcolor(red)"exact solutions"x=2±7exact solutions

Apr 14, 2018

x = 2+- sqrt(7)2±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)))⎜ ⎜b±(b)24(a)(c)2(a)⎟ ⎟

2/ ((-(-4)+-sqrt((-4)^2-4(1)(-3)))/(2(1)))⎜ ⎜(4)±(4)24(1)(3)2(1)⎟ ⎟

3/ ((4+-sqrt(16+12))/(2))(4±16+122)

4/ ((4+-2sqrt(7))/(2))(4±272) ( 2 cancel out)

5/ x = 2+-sqrt(7)2±7

Apr 14, 2018

x=2+sqrt7 or x=2-sqrt7x=2+7orx=27

Explanation:

Here,

x^2-4x-3=0x24x3=0

=>x^2-4x+4-7=0x24x+47=0

=>(x-2)^2=7=(sqrt7)^2(x2)2=7=(7)2

=>x-2=+-sqrt7x2=±7

=>x=2+-sqrt7x=2±7

OR

Comparing with quadratic equation,

ax^2+bx+c=0=>a=1,b=-4,c=-3ax2+bx+c=0a=1,b=4,c=3

triangle=b^2-4ac=(-4)^2-4(1)(-3)=b24ac=(4)24(1)(3)

=>triangle=16+12=28=4xx7=16+12=28=4×7

sqrt(triangle)=2sqrt7=27

So,

x=(-b+-sqrt(triangle))/(2a)x=b±2a

x=(4+-2sqrt7)/(2(1))x=4±272(1)

x=2+-sqrt7x=2±7