Saved Bookmarks
| 1. |
If a and b are positive integral numbers and each of the equations x2 + ax + 2b = 0, and x2 + 2bx + a=0 have real roots, then the smallest possible value of (a + b) is |
|
Answer» a and b are positive integral numbers and each of the equations x2 + ax + 2b = 0, and x² + 2bx + a=0 have REAL roots, To Find : smallest possible value of (a + b) isSolution:x² + ax + 2b = 0Real rootsif a² - 4(1)(2b) ≥ 0=> a² - 8b ≥ 0=> a² ≥ 8bx² + 2bx + a=0Real rootsif (2b)² - 4(1)(a) ≥ 0=> 4b² - 4a ≥ 0=> b² - a ≥ 0=> b² ≥ a => b ≥ √a a² ≥ 8b=> a² ≥ 8√aSquaring both sides=> a⁴ ≥ 64A=> a³ ≥ 64=> a³ ≥ 4³=> a ≥ 4b² ≥ a => b ≥ 2a ≥ 4 , b ≥ 2Smallest possible value of a + b = 4 + 2 = 6 smallest possible value of (a + b) is 6Learn More:Show that the roots of equation (XA)(xb) = abx^2; a,b belong R are ... brainly.in/question/21009259if alpha +BETA are the roots of equation 4x²-5x+2=0 find the equation ... brainly.in/question/8333453 |
|