The coefficient of x in x2+px+q=0 was taken as 17 in place of 13 and its roots were found to be -2 and -15 The roots of the original equation are:

Question

The coefficient of x in x2+px+q=0 was taken as 17 in place of 13 and its roots were found to be -2 and -15 The roots of the original equation are: (A) 3, 7 (B)  -3,7  (C)  -3,-7  (D)  3,10

Tardigrade Admin · Accepted Answer

The given equation is x2+17x+q=0 ∴ Product of roots =(-2)(-15)=q ⇒ q=30 ∴ Original equation is x2+13x+30=0 ⇒ x2+10x+3x+30=0 ⇒ (x+10)(x+3)=0 ⇒ x=-10,-3

Q. The coefficient of $x$ in $x^{2} + p x + q = 0$ was taken as 17 in place of 13 and its roots were found to be $- 2$ and $- 15$ The roots of the original equation are:

3, 7

$- 3, 7$

$- 3, - 7$

$3, 10$

$- 3, - 10$

The given equation is $x^{2} + 17 x + q = 0$ $∴$ Product of roots $= (- 2) (- 15) = q$ $\Rightarrow$ $q = 30$ $∴$ Original equation is $x^{2} + 13 x + 30 = 0$ $\Rightarrow$ $x^{2} + 10 x + 3 x + 30 = 0$ $\Rightarrow$ $(x + 10) (x + 3) = 0$ $\Rightarrow$ $x = - 10, - 3$

Q. The coefficient of x in x2+px+q=0 was taken as 17 in place of 13 and its roots were found to be −2 and −15 The roots of the original equation are:

3, 7

−3,7

−3,−7

3,10

−3,−10