What is x if #-4(x+2)^2=-20#?

3 Answers
Jul 21, 2018

Answer:

#" "#
Approximate values of #color(red)(x)# are:

#color(blue)(x~~0.236#

#color(blue)(x~~-4.236#

Explanation:

#" "#
How do we find the value of #color(red)(x# using

#color(red)([-4(x+2)^2]=(-20)#

#rArr (-4)(x^2+4x+4)=(-20)#

Divide both sides of the equation by #color(red)((-4)#

#rArr [(-4)(x^2+4x+4)]/((-4))=((-20))/((-4)#

#rArr [cancel (-4)(x^2+4x+4)]/(cancel ((-4)))=((-20))/((-4)#

#rArr x^2+4x+4=5#

Subtract #color(red)(5)# from both sides of the equation.

#rArr x^2+4x+4-5=5-5#

#rArr x^2+4x+4-5=cancel 5-cancel 5#

#rArr x^2+4x-1=0# ...Eqn.1

We now have a quadratic equation.

Use the quadratic formula to find the values of #color(red)(x#

Quadratic formula:

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

Using Eqn.1, we get

#a=1; b=4 and c= (-1)#

Substitute the values in the quadratic formula above:

#color(blue)(x=[-4+-sqrt(4^2-4*1*(-1))]/(2*1)#

#x=[-4+- sqrt(16+4)]/2#

#x=[-4+-sqrt(20)]/2#

#x=[-4+-sqrt(4*5)]/2#

#x=[-4+- sqrt(4)*sqrt(5)]/2#

#x=[-4+-2*sqrt(5)]/2#

#x=(-4/2)+-(2*sqrt(5))/2#

#x=(-2)+-(cancel 2*sqrt(5))/cancel 2#

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

#x=-2-sqrt(5)#, #x=-2+sqrt(5)#

Using a spreadsheet software or a calculator, we get

enter image source here

Hence, approximate values of #color(red)(x)# are:

#color(blue)(x~~0.236#

#color(blue)(x~~-4.236#

Hope it helps.

Jul 21, 2018

Answer:

#x=-2+sqrt5# and #-2-sqrt5#

Explanation:

Let's start by dividing both sides by #-4#. We now have

#(x+2)^2=5#

Let's take the square root of both sides to get

#x+2=sqrt5# and #x+2=-sqrt5#

Subtracting #2# from both sides of both equations, we get

#x=-2+sqrt5# and #-2-sqrt5#

Hope this helps!

Jul 21, 2018

Answer:

#x = -2 +- sqrt5#

Explanation:

#-4(x+2)^2 = -20#

First, divide both sides by #color(blue)(-4)#:
#(-4(x+2)^2)/color(blue)(-4) = (-20)/color(blue)(-4)#

#(x+2)^2 = 5#

Expand/simplify the left hand side:
#x^2 + 4x + 4 = 5#

Subtract #color(blue)5# from both sides:
#x^2 + 4x + 4 quadcolor(blue)(-quad5) = 5 quadcolor(blue)(-quad5)#

#x^2 + 4x - 1 = 0#

This is now in standard form, #ax^2 + bx + c#.

Use the quadratic formula #(-b +- sqrt(b^2 - 4ac))/(2a)# to find the value of #x#:
#x = (-4 +- sqrt(4^2 - 4(1)(-1)))/(2(1))#

#x = (-4 +- sqrt(16 + 4))/2#

#x = (-4 +- sqrt20)/2#

#x = (-4 +- 2sqrt5)/2#

#x = -2 +- sqrt5#

Hope this helps!