How do you solve # p^2+4p=-1#?

Question

4606 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-12T05:32:06

#x=-2+-sqrt3#

#color(blue)(p^2+4p=-1#

Add #1# both sides

#rarrp^2+4p+1=-1+1#

#color(purple)(rarrp^2+4p+1=0#

This is a Quadratic equation (in form #ax^2+bx+c=0#)

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

Remember that #aandb# are the #"coefficients"# of #p^2 and p # and #c# is the #"constant number"#

So,

#color(violet)((a,b,c)=(1,4,1))#

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

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

#rarrx=(-4+-sqrt(12))/(2)#

#rarrx=(-4+-sqrt(4*3))/(2)#

#rarrx=(-4+-2sqrt(3))/(2)#

Cancel

#rarrx=(-cancel4^2+-cancel2^1sqrt(3))/(cancel2^1)#

#color(green)(x=-2+-sqrt3#

Remember that the symbol #+-# means #"plus or minus"#

So,

It implies that,

#color(indigo)(x=-2+sqrt3,-2-sqrt3#

If you need more guidance for the Quadratic formula

Watch