If a root of the equation x^2 + px + 12 = 0 i | vedclass

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(C) Given that $4$ is a root of the equation $x^2 + px + 12 = 0$.Substituting $x = 4$ into the equation: $(4)^2 + p(4) + 12 = 0$.$16 + 4p + 12 = 0$.$4

Vedclass · Accepted Answer

$49/4$

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(C) Given that $4$ is a root of the equation $x^2 + px + 12 = 0$.Substituting $x = 4$ into the equation: $(4)^2 + p(4) + 12 = 0$.$16 + 4p + 12 = 0$.$4p + 28 = 0$.$4p = -28$,which gives $p = -7$.Now,consider the second equation $x^2 + px + q = 0$. Substituting $p = -7$,we get $x^2 - 7x + q = 0$.Since the roots of this equation are equal,the discriminant $D$ must be zero.$D = b^2 - 4ac = 0$.$(-7)^2 - 4(1)(q) = 0$.$49 - 4q = 0$.$4q = 49$.Therefore,$q = 49/4$.

If a root of the equation $x^2 + px + 12 = 0$ is $4$,while the roots of the equation $x^2 + px + q = 0$ are equal,then the value of $q$ will be

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