# Form a quadratic equation with integer coefficients and one of its roots as 3 + 4*sqrt 5.

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### 2 Answers

If one root of a quadratic equation is an irrational root 3+4sqrt5, then the other root of the quadratic equation should be 3-4sqrt5, as the irrational roots occur in pairs for a any polynomial whose coefficients are rational.

Further if x1 and x2 are the roots of a quadraatic equation, then we can write the quadratic equation f(x) = 0 inthe factor form, (x-x1)(x-x2) = 0.

Therefore , the required quadratic equation f(x) = 0 with roots x1 = 3+4sqrt5 and x2 = 3-4sqrt 5 is (x-(3+4sqrt5))(x-(3+4sqrt5)) = 0. We simplify the factor form into a polynomial form:

(x-3 +4sqrt)(x-3 - 4sqrt5) = 0.

(x-3)^2- (4sqrt5)^2 = 0 ,as (a+b)(a-b) = a^2 - b^2.

x^2-6x+9 -16*5 = 0.

x^2-6x+9-80 = 0.

x^2-6x-71 = 0 is the required quadratic equation with integral coefficients with roots 3+4sqrt5 and 3-4sqrt5.

Let the equation be ax^2 + bx + c =0

Now we have one of the roots as 3+ 4sqrt5, let the other root of the equation be R.

So we have (x- 3 - 4*sqrt 5)(x- R) = ax^2 + bx + c

=> x^2 – x*(3 - 4sqrt 5) – Rx + (3+4 sqrt 5)R = ax^2 + bx + c

=> x^2 – x*( 3- 4sqrt5 + R) + (3+4 sqrt 5)R = ax^2 + bx + c

Now if c is an integer R has to be equal to 3- 4 sqrt 5 as (3+4 sqrt 5) (3-4 sqrt 5) = 9 – 16*5 = 9 – 80 = -71

Therefore (x- 3 + 4*sqrt 5)(x- R) = ax^2 + bx + c

=> (x- 3 - 4*sqrt 5)(x- 3+ 4sqrt5) = ax^2 + bx + c

=> x^2 + x( -3-4sqrt5 -3 + 4 sqrt5) + 9 – 80 = ax^2 + bx + c

=> x^2 -6x -72 = ax^2 + bx + c

=> a =1 , b = -6 and c= -71

**Therefore the equation is x^2 -6x -71 = 0**