The coefficient of x in the equation x^2 + px | vedclass

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Question

(B) The incorrect equation used was $x^2 + 17x + q = 0$. Given that the roots of this incorrect equation are $-2$ and $-15$,we can find the value of $

Vedclass · Accepted Answer

$-3, -10$

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(B) The incorrect equation used was $x^2 + 17x + q = 0$. Given that the roots of this incorrect equation are $-2$ and $-15$,we can find the value of $q$ using the product of roots formula: $q = (-2) 	imes (-15) = 30$. Now,substitute $q = 30$ and the correct coefficient of $x$ (which is $13$) into the original equation: $x^2 + 13x + 30 = 0$. To find the roots,factorize the quadratic equation: $x^2 + 10x + 3x + 30 = 0$. $x(x + 10) + 3(x + 10) = 0$. $(x + 3)(x + 10) = 0$. Thus,the roots are $x = -3$ and $x = -10$.

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

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