Exercises What is the value of the discriminant and what is the number of real solutions for each equation? \( \begin{array}{lll}\text { 7. } x^{2}+x-42=0 & \text { 8. }-x^{2}+13 x-40=0 & \text { 9. } x^{2}+2 x+5=0 \\ \begin{array}{lll}\text { 10. } x^{2}=18 x-81 & \text { 11. }-x^{2}+7 x+44=0 & \text { 12. } \frac{1}{4} x^{2}-5 x+25=0\end{array} \\ \begin{array}{lll}\text { 13. } 2 x^{2}+7=5 x & \text { 14. } 4 x^{2}+25 x=21 & \text { 17. } \frac{1}{2} x^{2}+2 x+3=0\end{array} & \text { 18. } \frac{1}{6} x^{2}=2 x+18\end{array} \) Copyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved.
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To find the value of the discriminant and the number of real solutions for each given equation, we'll use the discriminant formula: \(D = b^2 - 4ac\). The number of real solutions can be determined as follows: if \(D > 0\), there are two real solutions; if \(D = 0\), there's one real solution; and if \(D < 0\), there are no real solutions. 1. For \(x^2 + x - 42 = 0\): - \(D = 1^2 - 4(1)(-42) = 1 + 168 = 169\) (two real solutions). 2. For \(-x^2 + 13x - 40 = 0\): - \(D = 13^2 - 4(-1)(-40) = 169 - 160 = 9\) (two real solutions). 3. For \(x^2 + 2x + 5 = 0\): - \(D = 2^2 - 4(1)(5) = 4 - 20 = -16\) (no real solutions). 4. For \(x^2 = 18x - 81\): - Rearranging gives \(x^2 - 18x + 81 = 0\), then \(D = (-18)^2 - 4(1)(81) = 324 - 324 = 0\) (one real solution). 5. For \(-x^2 + 7x + 44 = 0\): - \(D = 7^2 - 4(-1)(44) = 49 + 176 = 225\) (two real solutions). 6. For \(\frac{1}{4} x^2 - 5x + 25 = 0\): - \(D = (-5)^2 - 4(\frac{1}{4})(25) = 25 - 25 = 0\) (one real solution). 7. For \(2x^2 + 7 = 5x\): - Rearranging gives \(2x^2 - 5x + 7 = 0\), then \(D = (-5)^2 - 4(2)(7) = 25 - 56 = -31\) (no real solutions). 8. For \(4x^2 + 25x = 21\): - Rearranging gives \(4x^2 + 25x - 21 = 0\), then \(D = 25^2 - 4(4)(-21) = 625 + 336 = 961\) (two real solutions). 9. For \(\frac{1}{2}x^2 + 2x + 3 = 0\): - \(D = 2^2 - 4(\frac{1}{2})(3) = 4 - 6 = -2\) (no real solutions). 10. For \(\frac{1}{6}x^2 = 2x + 18\): - Rearranging gives \(\frac{1}{6}x^2 - 2x - 18 = 0\), then \(D = (-2)^2 - 4(\frac{1}{6})(-18) = 4 + 12 = 16\) (two real solutions). In summary, the equations yield the following solutions: 1. Two real solutions (169) 2. Two real solutions (9) 3. No real solutions (-16) 4. One real solution (0) 5. Two real solutions (225) 6. One real solution (0) 7. No real solutions (-31) 8. Two real solutions (961) 9. No real solutions (-2) 10. Two real solutions (16)
