Question

Two candidates attempt to solve a quadratic equation of the form x^2+px+q=0. One starts with a wrong value of p and finds the roots to be 2 and 6. The other starts with a wrong value of q and finds the roots to be 2 and – 9. Find the correct roots of the equation :

Answer 1

i think there is typing mistake.
in second case roots are -2 , 9
IN CASE ( 1 ).

here the values of p is wrong
So, 2 + 6 ≠ -b/a
8 ≠ -p/1
P ≠ -8

so he find the value of 2 × 6 = q/a correctly
12 = q/1
q = 12

IN CASE ( 2 )

here the value q is wrong
So, -2 × 9 ≠ q/a
-18 ≠ q/1
q ≠ -18

so, he find the value of (-2) + (9) = -p/a(correctly)
7 = -p/1
p = -7 [put in the Equation "x² + px + q = 0]

x² - 7x + 12 = 0
x² - 4x - 3x + 12 = 0
x(x - 4) - 3(x - 4) = 0
(x + 3)(x - 4) = 0
so, correct roots are ,
x = 3 , x = 4

Answer 2

Answer:

Hey .

Here is the answer.

If the roots are 2 and 6 then the quadratic equation will be

(x−2(x−6)=0⇒x2−6x−2x+12=0

x2−8x+12=0.

If the roots are 2 and −9 then the quadratic equation will be

(x−2(x+9)=0⇒x2+9x−2x−18=0

x2+7x−18=0

Now it is given thatp is wrong in the first quadratic equation and q is right and vice versa for the second quadratic equation

So, p=−8 is wrong and q=12 is right Similarly p=7 is right and q=−18 is wrong So taking the correct values of p and q our equation will become

x2+7x+12=0⇒x2+4x+3x+12=0

x=−3 and x=−4

Hope it helps you..