Question

Find the quadratic equation whose roots are $ 9 - \sqrt 5 ,9 + \sqrt 5 $
A. $ {x^2} - 18x - 56 = 0 $
B. $ {x^2} + 18x + 56 = 0 $
C. $ 4{x^2} - 72x - 56 = 0 $
D. $ {x^2} - 18x + 76 = 0 $

Answer 1

Hint: An equation which has two roots or of degree 2 is called a quadratic equation. If the roots of a quadratic equation are ‘a’ and ‘b’, then that quadratic equation in variable x will be in the form $ {x^2} - \left( {a + b} \right)x + ab = 0 $ . So add and multiply the given roots and substitute them in this general form accordingly to get the required equation.
Formula used:
 $ \left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2} $

Complete step-by-step answer:
We are given to find the quadratic equation whose roots are $ 9 - \sqrt 5 ,9 + \sqrt 5 $ .
Let $ 9 - \sqrt 5 $ be ‘a’ and $ 9 + \sqrt 5 $ be ‘b’.
When the roots of a quadratic equation are ‘a’ and ‘b’, then the equation will be $ {x^2} - \left( {a + b} \right)x + ab = 0 $
Where $ a + b $ is the sum of the roots and $ ab $ is the product of the roots of the quadratic equation
 $ a + b = \left( {9 - \sqrt 5 } \right) + \left( {9 + \sqrt 5 } \right) = 9 + 9 - \sqrt 5 + \sqrt 5 $
Cancelling the similar terms with different signs, we get $ a + b = 9 + 9 = 18 $
 $ ab = \left( {9 - \sqrt 5 } \right)\left( {9 + \sqrt 5 } \right) $
The RHS of the above equation is in the form $ \left( {x - y} \right)\left( {x + y} \right) $ which is equal to $ {x^2} - {y^2} $
Therefore, $ ab = {9^2} - {\left( {\sqrt 5 } \right)^2} $
 $ \Rightarrow ab = 81 - 5 = 76 $
On substituting the values of obtained $ a + b $ and $ ab $ in $ {x^2} - \left( {a + b} \right)x + ab = 0 $ , we get
 $ \Rightarrow {x^2} - \left( {18} \right)x + 76 = 0 $
Therefore the quadratic equation whose roots are $ 9 - \sqrt 5 ,9 + \sqrt 5 $ is $ {x^2} - 18x + 76 = 0 $
So, the correct answer is “Option D”.

Note: No. of roots of an equation is determined by its highest degree. If the highest degree is 3, then the equation will have 3 roots, if 4 then the equation will have 4 roots and so on. An equation can have both real roots and imaginary roots.