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.
Real Tutor Solution
Answer
Solution
Reviewed and approved by the UpStudy tutoring team
Beyond the Answer
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)