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# HELP!

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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
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1)

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

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Mar 17, 2019
#2
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2)

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

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Mar 17, 2019
#3
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3)

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

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Mar 17, 2019
#4
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4)

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

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Mar 17, 2019
#5
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5)

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

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Mar 17, 2019
#6
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$$(x-p)(x-q) = x^2+4x+6\\ (x-p)(x-q) = x^2+bx+c\\ \text{Comparing coefficients,} \\ (b,c) = (4,6)\\ \therefore b + c = 10$$

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Mar 17, 2019