(B) Step $1$: Solve equation $I$.$x - \sqrt{169} = 0$$x - 13 = 0$$x = 13$Step $2$: Solve equation $II$.$y^2 - 169 = 0$$y^2 = 169$$y = \pm \sqrt{169}$$

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(B) Step $1$: Solve equation $I$.$x - \sqrt{169} = 0$$x - 13 = 0$$x = 13$Step $2$: Solve equation $II$.$y^2 - 169 = 0$$y^2 = 169$$y = \pm \sqrt{169}$$y = 13$ or $y = -13$Step $3$: Compare $x$ and $y$.If $x = 13$ and $y = 13$,then $x = y$.If $x = 13$ and $y = -13$,then $x > y$.Combining these results,we get $x \ge y$.

Competitive Quantitative Aptitude Algebra QUADRATIC EQUATION

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Solve the given two equations and select the correct answer from the given options.
$I.$ $x - \sqrt{169} = 0$
$II.$ $y^2 - 169 = 0$

A
if $x > y$
B
if $x \ge y$
C
if $x < y$
D
if $x \le y$

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QUADRATIC EQUATION

If $\alpha$ and $\beta$ are the roots of the equation $x^{2}+3ax+c=0$ and if $\alpha^{2}+\beta^{2}=5$,find the value of $a$.

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If for a positive integer $n$,the quadratic equation $x(x + 1) + (x + 1)(x + 2) + \dots + (x + n - 1)(x + n) = 10n$ has two consecutive integral solutions,then $n$ is equal to:

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Find the value of $x^{3}-\frac{1}{x^{3}}$ when $x-\frac{1}{x}=a$.

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The least integral value $\alpha$ of $x$ such that $\frac{x - 5}{x^2 + 5x - 14} > 0$ satisfies:

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If the roots of the equation $x^2 + px + q = 0$ differ by $1$,then:

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Solve the given two equations and select the correct answer from the given options.$I.$ $x - \sqrt{169} = 0$$II.$ $y^2 - 169 = 0$

Similar Questions

If $\alpha$ and $\beta$ are the roots of the equation $x^{2}+3ax+c=0$ and if $\alpha^{2}+\beta^{2}=5$,find the value of $a$.

If for a positive integer $n$,the quadratic equation $x(x + 1) + (x + 1)(x + 2) + \dots + (x + n - 1)(x + n) = 10n$ has two consecutive integral solutions,then $n$ is equal to:

Find the value of $x^{3}-\frac{1}{x^{3}}$ when $x-\frac{1}{x}=a$.

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Solve the given two equations and select the correct answer from the given options.
$I.$ $x - \sqrt{169} = 0$
$II.$ $y^2 - 169 = 0$