# Let α and β be two solutions of the equation x^2-x+4=0. Then β/α+α/β=?

*print*Print*list*Cite

### 3 Answers

Let us start by solving for the roots of the given equation: x^2 -x+4 = 0

roots, `alpha` & `beta` = `(1 +- sqrt(1-4x4))/2 = (1 +- sqrt (1-16)) /2 = (1 +- sqrt(15)i)/2`

therefore, `alpha` = `(1+sqrt(15)i)/2` and `beta` = `(1-sqrt (15)i)/2`

Now, `beta/alpha + alpha/beta = (beta^2 + alpha^2)/(alpha beta)`

Here, `beta^2 + alpha^2 = (1-sqrt (15)i)/(2^2) + (1+sqrt(15)i)/(2^2) = (1-15 -2sqrt(15)i +1-15+2sqrt(15)i)/4 = -28/4 = -7`

and, `alpha beta = (1+sqrt(15)i)/2 x (1-sqrt(15)i)/2 = (1-15i^2)/(2^2) = 16/4 = 4`

Therefore, `beta/alpha + alpha/beta = -7/4`

Thus the answer is **-7/4**.

Hope that helps.

given equation is `x^2-x+4=0`

alpha and beta are roots roots

`(alpha)/(beta)+(beta)/(alpha)=((alpha)^2+(beta)^2)/(alpha beta)=((alpha+beta)^2-2alpha.beta)/(alpha.beta)`

`alpha+beta=1,alpha.beta=4`

substitute in the above expression then you will get answer as -7/4

Given the equation

the equation `x^2-x+4=0`

and the roots are `alpha` `,beta`

To find `(alpha/beta)+ (beta/alpha) `

`(alpha/beta)+ (beta/alpha) `` `= ` `` ``(alpha^2 +beta^2)/(alpha*beta) ` = `((alpha + beta)^2 -(2*alpha*beta))/(alpha*beta) ` = `(alpha+beta)^2/(alpha*beta) - 2 ` ------(1)

Now from the equation `x^2-x+4=0 `` ` is of the form `a*x^2 +b*x+c`

we know sum of the root of a quadratic equation is

`alpha+beta ` = `-b/a ` = `-(-1)/1 ` =1

and `alpha*beta ` = `c/a ` =`4/1 ` =4

substituting the values in expression we get

` `

` `

`(alpha/beta)+ (beta/alpha) `

`=(alpha+beta)^2 /(alpha*beta)-2`= `1/4 ` -2 = `-7/4 `

**Sources:**