Question

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

Answer 1

`x=-2+-sqrt3`

Explanation:

`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`)

Use the Quadratic formula

`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

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