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Question: If \[\\{ 8,2\\} \] are the roots of \[{x^2} + ax + \beta = 0\] and \[\\{ 3,3\\} \] are the roots of ...

If 8,2\\{ 8,2\\} are the roots of x2+ax+β=0{x^2} + ax + \beta = 0 and 3,3\\{ 3,3\\} are the roots of x2+αx+b=0{x^2} + \alpha x + b = 0, then the roots of the equation x2+ax+b=0{x^2} + ax + b = 0are:
(A). 1,−11, - 1
(B). −9,2 - 9,2
(C). −8,−2 - 8, - 2
(D). 9,19,1

Explanation

Solution

- Hint: Before finding the roots of the given equation, at first, we will find the value of a,ba,b.
Let us consider p,qp,q be the roots of a quadratic equation. Then, the equation can be written as,
x2−(p+q)x+pq=0{x^2} - (p + q)x + pq = 0

Complete step-by-step solution -
It is given that, 8,2\\{ 8,2\\} are the roots of x2+ax+β=0{x^2} + ax + \beta = 0. Also, given that, 3,3\\{ 3,3\\} are the roots of x2+αx+b=0{x^2} + \alpha x + b = 0.
We have to find the roots of the equation x2+ax+b=0{x^2} + ax + b = 0.
Before finding the roots of the given equation, at first, we will find the value of a,ba,b.
Let us consider, p,qp,q be the roots of a quadratic equation. Then, the equation can be written as,
x2−(p+q)x+pq=0{x^2} - (p + q)x + pq = 0
Since, 8,2\\{ 8,2\\} are the roots of a quadratic equation, the equation could be, x2−(8+2)x+8×2=0{x^2} - (8 + 2)x + 8 \times 2 = 0
Simplifying we get, the equation is, x2−10x+16=0{x^2} - 10x + 16 = 0
Comparing the equations, x2−10x+16=0{x^2} - 10x + 16 = 0 and x2+ax+β=0{x^2} + ax + \beta = 0 we get,
a=−10,β=16a = - 10,\beta = 16
Similarly,
Since, 3,3\\{ 3,3\\} are the roots of a quadratic equation, the equation could be, x2−(3+3)x+3×3=0{x^2} - (3 + 3)x + 3 \times 3 = 0
Simplifying we get, the equation is, x2−6x+9=0{x^2} - 6x + 9 = 0
Comparing the equations, x2−6x+9=0{x^2} - 6x + 9 = 0 and x2+αx+b=0{x^2} + \alpha x + b = 0 we get,
α=−6,b=9\alpha = - 6,b = 9
Substitute the value of a=−10,b=9a = - 10,b = 9 in the equation x2+ax+b=0{x^2} + ax + b = 0we get,
The equation as: x2−10x+9=0{x^2} - 10x + 9 = 0
Now, we will apply the middle term factor method to find out the roots.
So, the equation can be written as,
x2−(9+1)x+9=0{x^2} - (9 + 1)x + 9 = 0
Simplifying we get,
x2−9x−x+9=0{x^2} - 9x - x + 9 = 0
Simplifying again we get,
(x−9)(x−1)=0(x - 9)(x - 1) = 0
Hence, the roots are 9,19,1

Hence, the correct option is (D) 9,19,1

Note:
To find the roots of the quadratic equation we can apply Sreedhar Acharya’s formula instead of middle term factorization.
It states that, if ax2+bx+c=0a{x^2} + bx + c = 0 be a quadratic equation then its roots are,
x=−b±b2−4ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}
We will find the same answer as above.