+21 hii, please help, thx!!
QUESTION 1-
The quadratic equation \(2x^2 + bx + 18 = 0 \) has a double root. Find all possible values of \(b.\)
QUESTION 2-
Let \(r \) and \(s\) be the roots of \(x^2 - 6x + 2 = 0\) Find \((r - s)^2.\)
QUESTION 3-
The roots of \(3x^2 - 4x + 15 = 0\) are the same as the roots of \(x^2 + bx + c = 0,\) for some constants \(b\) and \(c\) Find the ordered pair \((b,c)\)
+505 1. Guest was almost right, but he/she forgot to multiply by 2 when squaring the binomial. So the answer is 12.
2. \((r-s)^2 = r^2-2rs+s^2\). By Vieta's formulas, \(r+s = \frac{-6}{-1} = 6\), and \(rs = \frac{2}{1} = 2\) so \((r+s)^2= r^2+2rs+s^2 = 6^2=36\). Now just subtract \(4rs=8\) from 36 to get what we want, so the final answer is 28.
3. Divide \(3x^2-4x+15=0\) by 3 to get \(x^2 - \frac{4}{3}x + 5 = 0\). Therefore, the ordered pair is \((\frac{-4}{3}, 5)\)
