Question

If P and Q tried to solve the quadratic equation x^2+bx+c=0, P by mistake took the wrong value of b and found the roots to be 12,2. Q did a similar mistake by taking the wrong side of c and found the roots to be 2,8. Find the actual roots of the equation

Answer 1

HELLO DEAR,

A to Q,

condition --- ( 1 )

P by mistake took the wrong value of b and found the roots to be 12 , 2 .

roots are 12 & 2

=> (x - 12)(x - 2)

=> x² - 2x - 12x + 24 = 0

=> x² - 14x + 24 = 0

(as mistake is only in x term so, c = 24)

condition --- ( 2 )

Q did a similar mistake by taking the wrong side of c and found the roots to be 2 , 8.

roots are 2 & 8

=> (x - 2)(x - 8) = 0

=> x² - 8x - 2x + 16 = 0

=> x² - 10x + 16 = 0

(as mistake is only in constant term so, b = -10)

therefore, the correct equation is

x² - 10x + 24 = 0

=> x² - 6x - 4x + 24 = 0

=> x(x - 6) - 4(x - 6) = 0

=> (x - 4)(x - 6) = 0

=> x = 4 & x = 6

I HOPE IT'S HELP YOU DEAR,

THANKS