+0

# halp meh

0
505
7
+21

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
-3

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

Jan 26, 2021
#2
+21
0

it's not right :/ but thx for trying

esgheasgyjjjjjuhga  Jan 26, 2021
#3
+505
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
+21
-1

1 is still wrong

esgheasgyjjjjjuhga  Jan 27, 2021
#6
+505
0

You're right, I forgot about -12

textot  Jan 27, 2021
#5
-2

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
+129465
+2

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

Jan 27, 2021