The product of real roots of the equation |x| | vedclass

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Question

(A) Given equation is $|x|^{6/5} - 26|x|^{3/5} - 27 = 0$. Let $|x|^{3/5} = t$. Since $|x|^{3/5} \ge 0$,we have $t \ge 0$. The equation becomes $t^2 -

Vedclass · Accepted Answer

$-3^{10}$

Vedclass · Answer

(A) Given equation is $|x|^{6/5} - 26|x|^{3/5} - 27 = 0$. Let $|x|^{3/5} = t$. Since $|x|^{3/5} \ge 0$,we have $t \ge 0$. The equation becomes $t^2 - 26t - 27 = 0$. Factoring the quadratic: $(t - 27)(t + 1) = 0$. This gives $t = 27$ or $t = -1$. Since $t \ge 0$,we discard $t = -1$. Thus,$|x|^{3/5} = 27$. $|x|^{3/5} = 3^3$. Raising both sides to the power of $5/3$: $|x| = (3^3)^{5/3} = 3^5$. Therefore,$x = 3^5$ or $x = -3^5$. The product of the real roots is $(3^5) 	imes (-3^5) = -3^{10}$.

The product of real roots of the equation $|x|^{6/5} - 26|x|^{3/5} - 27 = 0$ is:

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