(A) Given equation is $|x|^{6/5} - 26|x|^{3/5} - 27 = 0$. Let $|x|^{3/5} = t$. Then the equation becomes $t^2 - 26t - 27 = 0$. Factoring the quadratic

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

Class 11 Mathematics 2 Complex Numbers and Quadratic Equations: Quadratic Equations Solution of quadratic equations and Nature of roots

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Find the product of all real roots of the equation $|x|^{6/5} - 26|x|^{3/5} - 27 = 0$.

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A
$-3^{10}$
B
$-3^{12}$
C
$-3^{12/5}$
D
$-3^{21/5}$

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2 Complex Numbers and Quadratic Equations: Quadratic Equations

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Solution of quadratic equations and Nature of roots

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Find the product of all real roots of the equation $|x|^{6/5} - 26|x|^{3/5} - 27 = 0$.

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