Question

Question #75433

Answer 1

Any cubic equations with three roots (e.g. `x^3 - 6x^2 + 4x + 12`)

Explanation:

Step One: Find the number of solutions in the equation

I'm not sure if you have learnt about discriminant yet but I'll explain.

The discriminant basically tells you how many roots there are in a quadratic equation.

The discriminant is `b^2 - 4ac` for the quadratic equation `ax^2 + bx + c = 0`

If the discriminant is bigger than `0`, there are two roots.
If the discriminant is equal to `0`, there are one root.
If the discriminant is smaller than `0`, there are no roots.

Therefore if you look at the equation, `x^2 + 10x + 22 = 0`, the discriminant is:

`b^2 - 4ac = 10^2 - 4\times1\times22 = 12`.
Therefore, this equation has two roots.

If you don't want to use this method, you can just solve the equation and find it out manually.

Step two: Form the required equation

If we want an equation that has one more than two roots (i.e. three roots), we need to form a cubic equation.

A cubic equation can have up to three roots can its highest power of `x` is `3`.

Therefore, any cubic functions which have three solutions is the answer.

For example, `x^3 - 6x^2 + 4x + 12`.

Hope that makes sense!

Answer 2

`x^2 + 8x + 13 = 0`

Explanation:

The root form of are quadratic with roots `r_1` and `r_2` is:

`(x-r_1)(x-r_2) = 0`

Perform the multiplication:

`x^2 -(r_1+r_2)x + r_1r_2 = 0`

Matching the coefficients of the terms of the general root form with the coefficients of the given equation, `x^2 + 10x + 22 = 0`, we can write:

`r_1+r_2 = -10" [1]"`

`r_1r_2 = 22" [2]"`

Add 1 to the roots:

`(x-(r_1+1)(x-(r_2+1)) = 0`

Perform the multiplication:

`x^2 -((r_1+1)+(r_2+1))x + (r_1+1)(r_2+1) = 0" [3]"`

Expanding the middle coefficient:

`(r_1 +1) + (r_2+1) = r_1 + r_1 + 2`

Substitute the right side of equation [1]:

`(r_1 +1) + (r_2+1) = -10 + 2`

`(r_1 +1) + (r_2+1) = -8`

Expanding the constant term:

`(r_1+1)(r_2+1) = r_1r_2+ r_1 + r_2 + 1`

Substitute in equation [1] and equation [2]:

`(r_1+1)(r_2+1) = 22-10 + 1`

`(r_1+1)(r_2+1) = 13`

Substitute these value into equation [3]:

`x^2 + 8x + 13 = 0`

The roots of `x^2 + 10x + 22 = 0` are `-5+-sqrt3`

The roots of `x^2 + 8x + 13 = 0` are `-4+-sqrt3`

Answer 3

The quadratic equation is `x^2+8x+13=0`

Explanation:

Let `alpha and beta` are the roots of the equation

`x^2+10x+22=0 ; a= 1 , b=10 ,c =22`. Then,

`alpha+beta = -b/a= -10/1=-10 and alpha*beta = c/a= 22/1=22`

Then `(alpha+1) and (beta+1)` are the roots of the new equation,

`a_1x^2+b_1x+c_1=0;a_1=1`

`(alpha+1)+(beta+1) = alpha+beta+2 =-10+2=-8`

`:. b_1=-(-8)=8` and (alpha+1)*(beta+1)#

`= alpha*beta+alpha+beta+1=22-10+1=13`

`:. c_1=13`. So the quadratic equation is `x^2+8x+13=0` [Ans]