2. Between which two consecutive integers does \( \sqrt{138} \) lie? .3. Given \( \mathrm{q}=\sqrt{b^{2}-4 a c} \). If \( \mathrm{a}=-1 ; \mathrm{b}=-2 \) and \( \mathrm{c}=4 \). Determine between which two Consecutive integers does q lie.
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To find where \( \sqrt{138} \) lies, let's consider the perfect squares close to 138. We know that \( 12^2 = 144 \) and \( 11^2 = 121 \). Since \( 121 < 138 < 144 \), we can conclude that \( \sqrt{138} \) lies between 11 and 12. Now, for the expression \( \mathrm{q} = \sqrt{b^2 - 4ac} \) with \( \mathrm{a} = -1 \), \( \mathrm{b} = -2 \), and \( \mathrm{c} = 4 \), we first calculate \( b^2 - 4ac \). Calculating this gives: \[ b^2 = (-2)^2 = 4 \] \[ 4ac = 4 \times (-1) \times 4 = -16 \] \[ b^2 - 4ac = 4 - (-16) = 4 + 16 = 20 \] So, we find \( \mathrm{q} = \sqrt{20} \). Since \( 4^2 = 16 \) and \( 5^2 = 25 \), it shows that \( \sqrt{20} \) lies between 4 and 5.