If a and b are the roots of x² - x - 12 = 0, and a > b, then the quadratic equation in x whose roots are (2a- 1) and (2b + 1) is:

Question

Question

This question was previously asked in

RRB Group D 25 Aug 2022 Shift 2 Official Paper

Answer 1

Given:

Roots of equation x2 - x - 12 = 0, = a and b

Given roots: (2a - 1) and (2b + 1)

Concept used:

In equation ax2 + bx + c = 0,

Sum of roots = \(-b \over a\)

Product of roots = \(c \over a\)

Difference of roots = \({ \sqrt{b^2-4ac} \over a}\)

Calculation:

From equation x2 - x - 12 = 0, we get

\(a + b = 1\\ ab = -12\\ a -b= { \sqrt{1^2-4(-12)} \over 1}\\ = { \sqrt{1+48}}\\= { \sqrt{49}}=7\)

Substituting this value in the below equations:

Given roots: (2a - 1) and (2b + 1)

\(\text{Sum of roots }=2a - 1 + 2b + 1 = 2a + 2b = 2(a + b) = 2(1) = 2\)

\(\text{Product of roots} = (2a - 1) × (2b + 1) = 4ab + 2a - 2b - 1 = 4(-12) + 2(a-b) - 1 = - 48 + 2(7) -1 = - 35\)

Thus, the quadratic equation formed will be:

x2 - (sum of roots) x + product of roots = 0

⇒ x2 - 2x + (-35) = 0

⇒ x2 - 2x - 35 = 0

Hence, the quadratic equation whose roots are (2a - 1) and (2b + 1) is x2 - 2x - 35 = 0.