Question

If a and b are the zeroes of \[{x^2} - 2x + 3 = 0\], find the polynomial whose zeroes are a+2, b+2.

Answer 1

Hint: Zeroes means the same as roots. Use the relation between the coefficients of the quadratic equation and the sum and product of roots, to establish relations between the given roots. Use these formed relations to form a polynomial.

Complete step-by-step answer:
Given quadratic equation
\[ \Rightarrow {x^2} - 2x + 3 = 0\]
Now it is given that a and b are the zeroes of the equation, and we know zeroes are nothing but the roots of the equation.
Now as we know that the sum of the roots of the quadratic equation is the ratio of negative times the coefficient of x to the coefficient of x2.
$ \Rightarrow a + b = \dfrac{{ - \left( { - 2} \right)}}{1} = 2$……………………. (1)
And the product of roots is the ratio of constant term to the coefficient of x2.
$ \Rightarrow ab = \dfrac{3}{1} = 3$………………………………. (2)
Now we have to find the quadratic equation whose roots are (a + 2) and (b + 2).
So the sum (S) of the roots is (a + 2 + b +2) = a + b +4
From equation (1) we have,
$ \Rightarrow S = a + b + 4 = 2 + 4 = 6$
And the product (P) of the roots = (a + 2)(b + 2)
$ \Rightarrow P = ab + 2\left( {a + b} \right) + 4$
Now from equation (1) and (2) we have,
$ \Rightarrow P = 3 + 2\left( 2 \right) + 4 = 3 + 4 + 4 = 11$
So the general quadratic equation is
$ \Rightarrow {x^2} - \left( {{\text{sum of roots}}} \right)x + \left( {{\text{product of roots}}} \right) = 0$
So substitute the value of S and P in the above equation we have,
$ \Rightarrow {x^2} - 6x + 11 = 0$
So this is the required quadratic equation.

Note: Whenever we face such type of problems the key concept is to have a good grasp over the direct formula to obtain the polynomial when sum and products of its roots can be evaluated which is ${x^2} - \left( {{\text{sum of roots}}} \right)x + \left( {{\text{product of roots}}} \right) = 0$. This helps in solving problems of this kind and helps saving a lot of time.