Two candidates attempt to solve a quadratic equation x2 + px + q = 0. One start with a wrong value of p and gets 2, 6 as its roots and other start with a wrong value of q and obtained roots 2, - 9. The correct roots are

Question

Two candidates attempt to solve a quadratic equation x2 + px + q = 0. One start with a wrong value of p and gets 2, 6 as its roots and other start with a wrong value of q and obtained roots 2, - 9. The correct roots are (A) - 3 , - 4 (B) 3, 4 (C) - 3 ,4 (D) - 4 ,3

Tardigrade Admin · Accepted Answer

Consider x2+p x+q=0, by assuming α, β as correct roots ∴ α β=q=2 × 6=12 the wrong value of p effects only the sum of the roots and not the product, we need p. Now, again for second candidate, assume value of p is correct. ∴ α+β=p=2+(-9)=-7 ∴ the correct equation is x2-( Sum of roots ) x+ product of roots =0 or x2+7 x+12= 0 ⇒ x=-3,-4

Q. Two candidates attempt to solve a quadratic equation $x^{2} + p x + q = 0$ . One start with a wrong value of $p$ and gets $2, 6$ as its roots and other start with a wrong value of $q$ and obtained roots $2, - 9$ . The correct roots are

$- 3, - 4$

$3, 4$

$- 3, 4$

$- 4, 3$

Q. Two candidates attempt to solve a quadratic equation x2+px+q=0. One start with a wrong value of p and gets 2,6 as its roots and other start with a wrong value of q and obtained roots 2,−9. The correct roots are

−3,−4

3,4

−3,4

−4,3

Q. Two candidates attempt to solve a quadratic equation $x^{2} + p x + q = 0$ . One start with a wrong value of $p$ and gets $2, 6$ as its roots and other start with a wrong value of $q$ and obtained roots $2, - 9$ . The correct roots are

$- 3, - 4$

$3, 4$

$- 3, 4$

$- 4, 3$