Solve the given two equations and select the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) For equation $I$: $\sqrt{x} = \frac{(18)^{15/2}}{x^2}$Multiplying both sides by $x^2$,we get $x^{1/2} \cdot x^2 = (18)^{15/2}$$x^{5/2} = (18)^{15/

Vedclass · Accepted Answer

if $x < y$

Vedclass · Answer

(C) For equation $I$: $\sqrt{x} = \frac{(18)^{15/2}}{x^2}$Multiplying both sides by $x^2$,we get $x^{1/2} \cdot x^2 = (18)^{15/2}$$x^{5/2} = (18)^{15/2}$Raising both sides to the power of $2/5$,we get $x = (18)^{(15/2) \cdot (2/5)} = 18^3 = 5832$.For equation $II$: $\sqrt{y} = \frac{(19)^{9/2}}{y}$Multiplying both sides by $y$,we get $y^{1/2} \cdot y = (19)^{9/2}$$y^{3/2} = (19)^{9/2}$Raising both sides to the power of $2/3$,we get $y = (19)^{(9/2) \cdot (2/3)} = 19^3 = 6859$.Comparing the values,$x = 5832$ and $y = 6859$.Since $5832 < 6859$,therefore $x < y$.

Solve the given two equations and select the correct answer from the given options.
$I.$ $\sqrt{x} - \frac{(18)^{15/2}}{x^2} = 0$
$II.$ $\sqrt{y} = \frac{(19)^{9/2}}{y}$

Similar Questions

If $\alpha$ and $\beta$ are the roots of the equation $ax^2 + bx + c = 0$ ($a \ne 0$; $a, b, c$ being distinct),then $(1 + \alpha + \alpha^2)(1 + \beta + \beta^2) = $

If $x = \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$ and $y = \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$,then the value of $\frac{x^{2}+xy+y^{2}}{x^{2}-xy+y^{2}} = ?$

The roots of the equation $a^{2} x^{2}-3 a b x+2 b^{2}=0$ are

The roots of the equation $4^x - 3 \cdot 2^{x+2} + 32 = 0$ are

Solve the given two equations and select the correct answer from the given options.
$I.$ $(17)^{2} + 144 \div 18 = x$
$II.$ $(26)^{2} - 18 \times 21 = y$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Solve the given two equations and select the correct answer from the given options.$I.$ $\sqrt{x} - \frac{(18)^{15/2}}{x^2} = 0$$II.$ $\sqrt{y} = \frac{(19)^{9/2}}{y}$