If the roots of the equation x^2 + a^2 = 8x + | vedclass

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Question

(B) The given equation is $x^2 - 8x + (a^2 - 6a) = 0$.For the roots to be real,the discriminant $D$ must be greater than or equal to zero $(D \ge 0)$.

Vedclass · Accepted Answer

$a \in [-2, 8]$

Vedclass · Answer

(B) The given equation is $x^2 - 8x + (a^2 - 6a) = 0$.For the roots to be real,the discriminant $D$ must be greater than or equal to zero $(D \ge 0)$.Here,$A = 1$,$B = -8$,and $C = a^2 - 6a$.$D = B^2 - 4AC = (-8)^2 - 4(1)(a^2 - 6a) \ge 0$.$64 - 4a^2 + 24a \ge 0$.Dividing by $-4$ (and reversing the inequality sign): $a^2 - 6a - 16 \le 0$.Factoring the quadratic: $(a - 8)(a + 2) \le 0$.This inequality holds when $a$ lies between the roots $-2$ and $8$.Therefore,$a \in [-2, 8]$.

If the roots of the equation $x^2 + a^2 = 8x + 6a$ are real,then:

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