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

(D) For equation $I$:$\frac{15}{\sqrt{x}}-\frac{9}{\sqrt{x}}=\sqrt{x}$$\frac{6}{\sqrt{x}}=\sqrt{x}$$6 = \sqrt{x} \cdot \sqrt{x}$$x = 6$For equation $I

Vedclass · Accepted Answer

if $x = y$ or relationship between $x$ and $y$ cannot be established.

Vedclass · Answer

(D) For equation $I$:$\frac{15}{\sqrt{x}}-\frac{9}{\sqrt{x}}=\sqrt{x}$$\frac{6}{\sqrt{x}}=\sqrt{x}$$6 = \sqrt{x} \cdot \sqrt{x}$$x = 6$For equation $II$:$y^{10} - (36)^{5} = 0$$y^{10} = (6^2)^5$$y^{10} = 6^{10}$Since the exponent is even,$y = 6$ or $y = -6$.Comparing $x$ and $y$:If $x = 6$ and $y = 6$,then $x = y$.If $x = 6$ and $y = -6$,then $x > y$.Since both conditions are possible,the relationship between $x$ and $y$ cannot be established.

Solve the given two equations and select the correct answer from the given options.
$I.$ $\frac{15}{\sqrt{x}}-\frac{9}{\sqrt{x}}=(x)^{\frac{1}{2}}$
$II.$ $y^{10}-(36)^{5}=0$

Similar Questions

Solve the given two equations and select the correct answer from the given options.
$I.$ $8x^2 + 6x = 5$
$II.$ $12y^2 - 22y + 8 = 0$

Determine $k$ such that the quadratic equation $x^{2}-2(1+3k)x+7(3+2k)=0$ has equal roots.

If the difference between the roots of the equation $x^2 + ax + b = 0$ is equal to the difference between the roots of the equation $x^2 + bx + a = 0$ $(a \ne b)$,then:

Given that $x^{3}+y^{3}=72$ and $xy=6$ with $x>y$,then the value of $x-y$ is:

The roots of the equation $x^{2}+px+q=0$ are equal if:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Solve the given two equations and select the correct answer from the given options.$I.$ $\frac{15}{\sqrt{x}}-\frac{9}{\sqrt{x}}=(x)^{\frac{1}{2}}$$II.$ $y^{10}-(36)^{5}=0$