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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 \((b,c)\)

 Jan 26, 2021
 #1
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-2

1. Complete the square: 2(x + 3)^2 = 2x^2 + 6x + 18 ==> b = 6.

 Jan 26, 2021
 #2
avatar+11 
0

it's not right :/ but thx for trying

esgheasgyjjjjjuhga  Jan 26, 2021
 #3
avatar+165 
0

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 \((\frac{-4}{3}, 5)\)

 Jan 27, 2021
 #4
avatar+11 
-1

1 is still wrong

esgheasgyjjjjjuhga  Jan 27, 2021
 #6
avatar+165 
0

You're right, I forgot about -12

textot  Jan 27, 2021
 #5
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-1

Question 2: The roots of x^2 - 6x + 2 = 0 are x = 3 - 2*sqrt(2) and 3 + 2*sqrt(2), so (r - s)^2 = (4*sqrt(2))^2 = 32.

 Jan 27, 2021
 #7
avatar+116124 
+1

1.  If  2x^2  + bx + 18  =  0   has a double root......then the discriminant   =  0 

 

So

 

b^2  - 4 * 2 * 18  = 0

 

b^2  -  144   = 0

 

b^2   = 144

 

b  =12   or  b =  -12

 

 

cool cool cool

 Jan 27, 2021

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