A) If x>y |
B) \( \Large If \ x\ge y \) |
C) lfx |
D) If x = y or relation cannot be established |
D) If x = y or relation cannot be established |
\( \Large 225x^{2}-4=0 \)
=> \( \Large 225x^{2}=4 => x^{2}=\frac{4}{225} \)
Therefore, \( \Large x=\sqrt{\frac{4}{225}}=\pm \frac{2}{15}, \ i.e., \ \frac{2}{15} \ and -\frac{2}{15} \)
and \( \Large \sqrt{225y}+2=0 \ or \sqrt{225y}=-2 \)
On squaring both sides, we get
\( \Large \sqrt{\left(225y\right)^{2}} = \left(-2\right)^{2} \)
So the relationship cannot be established because \( \Large y = \frac{4}{225} \)
lies between \( \Large \frac{2}{15} and -\frac{2}{15} \)
1). \( \Large \frac{4}{\sqrt{x}}+\frac{7}{\sqrt{x}}=\sqrt{x;} \) \( \Large y^{2}-\frac{ \left(11\right)^{\frac{5}{2}} }{\sqrt{y}} =0 \)
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2). \( \Large x^{2}-365=364; y-\sqrt{324}=\sqrt{81} \)
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3). \( \Large 3x^{2}+8x+4=0; 4y^{2}-19y+12=0 \)
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4). \( \Large x^{2}-x-12=0; y^{2}+5y+6=0 \)
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5). \( \Large x^{2}-8x+15=0; y^{2}-3y+2=0 \)
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6). \( \Large x^{2}-32=112; y-\sqrt{169}=0 \)
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7). \( \Large x-\sqrt{121}=0; y^{2}-121=0 \)
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8). \( \Large x^{2}-16=0; y^{2}-9y+20=0 \)
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9). If \( \Large x^{2} = 6 + \sqrt{6+\sqrt{6 + \sqrt{6+.... ...... \infty}}} \) then what is one of the values of x equal to?
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10). The sum of a number and its reciprocal is \( \Large \frac{10}{3} \), then the numbers are
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