We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
132
6
avatar

1. Let d and e denote the solutions of 3x^2+10x-25=0. Find (d-e)^2.

2. Let a and b be the solutions of the quadratic equation \(2x^2 - 8x + 7 = 0\). Find \(\frac{1}{2a} + \frac{1}{2b}.\)
3. Let s and t be the solutions of the quadratic \(4x^2 + 9x - 6 = 0\). Find \(\frac st + \frac ts.\)
4. Let m and n be the solutions of the quadratic equation \(4x^2 + 5x + 3 = 0\). Find \((m + 7)(n + 7)\).

5. Find x-y if the pair (x,y) satisfies \(x^4=y^4+18\sqrt3\), \(x^2+y^2=6\), and \(x+y=3\).

6. p and q are the solutions to the quadratic equation \(x^2 + 4x + 6 = 0\)
    p and q are the solutions to the quadratic equation \(x^2 + bx + c = 0\)
    Find b + c

 Mar 16, 2019
 #1
avatar+7709 
0

1)

\(d+e = \dfrac{-10}{3}\\ de = \dfrac{-25}{3}\\ (d-e)^2 = (d+e)^2 - 4de = \left(\dfrac{-10}{3}\right)^2 - 4\left(\dfrac{-25}{3}\right) = \dfrac{400}{9}\)

.
 Mar 17, 2019
 #2
avatar+7709 
0

2) 

\(a+b = -\dfrac{-8}{2} = 4\\ ab = \dfrac{7}{2}\\ \dfrac{1}{2a} + \dfrac{1}{2b} = \dfrac{1}{2}\left(\dfrac{a+b}{ab}\right) = \dfrac{1}{2}\left(\dfrac{4}{\frac{7}{2}}\right)=\dfrac{4}{7}\)

.
 Mar 17, 2019
 #3
avatar+7709 
0

3)

\(s+t = -\dfrac{9}{4}\\ st = \dfrac{-6}{4} = -\dfrac{3}{2}\\ \dfrac{s}{t} + \dfrac{t}{s} = \dfrac{(s+t)^2 - 2st}{st} = \dfrac{\left(\frac{-9}{4}\right)^2 -2\left(\frac{-3}{2}\right)}{\frac{-3}{2}} =\dfrac{-43}{8}\)

.
 Mar 17, 2019
 #4
avatar+7709 
0

4)

\(m+n = \dfrac{-5}{4}\\ mn = \dfrac{3}{4}\\ (m+7)(n+7) = mn + 7(m+n) + 49 = \dfrac{3}{4}+7\left(\dfrac{-5}{4}\right) + 49 = 41\)

.
 Mar 17, 2019
 #5
avatar+7709 
0

5)

\(\begin{cases} x+y=3\\ x^2+y^2=6\\ x^4=y^4+18\sqrt3 \end{cases}\\ (x + y)^2 = 9\\ x^2 + y^2 + 2xy = 9\\ 2xy = 3\\ x = \dfrac{3}{2y}\\ \dfrac{3}{2y} + y = 3\\ 2y^2 - 6y + 3 = 0\\ y = \dfrac{6\pm2\sqrt{3}}{4} = \dfrac{3\pm\sqrt{3}}{2}\\ \text{When } y = \dfrac{3+\sqrt3}{2}\text{,}\\ x = \dfrac{3-\sqrt3}{2}\\ \text{When } y = \dfrac{3-\sqrt3}{2}\text{,}\\ x = \dfrac{3+\sqrt3}{2}\\ \therefore (x,y) = \left(\dfrac{3+\sqrt3}{2},\dfrac{3-\sqrt3}{2}\right) \text{ OR }\left(\dfrac{3-\sqrt3}{2},\dfrac{3+\sqrt3}{2}\right)\)

.
 Mar 17, 2019
 #6
avatar+7709 
0

\((x-p)(x-q) = x^2+4x+6\\ (x-p)(x-q) = x^2+bx+c\\ \text{Comparing coefficients,} \\ (b,c) = (4,6)\\ \therefore b + c = 10\)

.
 Mar 17, 2019

15 Online Users

avatar
avatar