(D) Given the equation $x^{2/3} - 7x^{1/3} + 10 = 0.$This can be rewritten as $(x^{1/3})^2 - 7(x^{1/3}) + 10 = 0.$Let $a = x^{1/3}.$ Substituting this

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(D) Given the equation $x^{2/3} - 7x^{1/3} + 10 = 0.$This can be rewritten as $(x^{1/3})^2 - 7(x^{1/3}) + 10 = 0.$Let $a = x^{1/3}.$ Substituting this into the equation,we get $a^2 - 7a + 10 = 0.$Factoring the quadratic equation: $(a - 5)(a - 2) = 0.$This gives $a = 5$ or $a = 2.$Since $a = x^{1/3},$ we have $x = a^3.$For $a = 5,$ $x = 5^3 = 125.$For $a = 2,$ $x = 2^3 = 8.$Therefore,$x = 125, 8.$

Class 11 Mathematics 4-2.Quadratic Equations and Inequations Solution of quadratic equations and Nature of roots

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If $x^{2/3} - 7x^{1/3} + 10 = 0,$ then $x = \dots$

A
$125$
B
$8$
C
$\phi$
D
$125, 8$

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4-2.Quadratic Equations and Inequations

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Solution of quadratic equations and Nature of roots

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Let $p, q, r \in \mathbb{R}$ and $r > p > 0.$ If the quadratic equation $px^2 + qx + r = 0$ has two complex roots $\alpha$ and $\beta,$ then $|\alpha| + |\beta|$ is:

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If the difference between the roots of the equations $x^2 - bx + c = 0$ and $x^2 - cx + b = 0$ is the same,then $b + c = \dots$

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The smallest value of the constant $m > 0$ for which $f(x) = 9mx - 1 + \frac{1}{x} \geq 0$ for all $x > 0$ is:

The equation $x^4-x^3-6x^2+4x+8=0$ has two equal roots. If $\alpha$ and $\beta$ are the other two roots of this equation,then $\alpha^2+\beta^2=$

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