+0  
 
+1
22
10
avatar+77 

1. Fill in the blanks to make a quadratic whose roots are 5 and -3.

 

\(x^2+\boxed{\color{white}ABC}x+\boxed{\color{white}ABC} \)

 

2. Let \(a\) and \(b\) be the roots of the quadratic  Find the quadratic whose roots are \(a^2\) and \(b^2\).

 

\(x^2+\boxed{\color{white}ABC}x+\boxed{\color{white}ABC} \)

 

3. Let \(a\) and \(b\) be the roots of the quadratic \(2x^2 - 8x + 7 = 0\). Compute \(a^3 + b^3.\)

 

4. Fill in the blanks to make the equation true.

 

\((\boxed{\color{white}ABC}x+\boxed{\color{white}ABC})(\boxed{\color{white}ABC}x+\boxed{\color{white}ABC})=\boxed{\color{white}ABC}x^2+\boxed{\color{white}ABC}x+\boxed{\color{white}ABC}\)

 

Options:

  • -35
  • -8
  • -5
  • 1
  • 3
  • 4
  • 6
  • 7
  • 12
 Feb 1, 2024
 #1
avatar+129895 
+2

1.  (x - 5) (x + 3)  =  0

      x^2   - 2 x + - 15  = 0

 

3.    2x^2  -8x  + 7  =   0

 

Product of roots  = ab = 7/2

2ab  = 7       (1)

 

Sum of roots  =  a + b = - (-8) / 2 = 4

Square both sides

a^2 + 2ab + b^2 = 16      (2)

Sub (1) into (2)

a^2 + 7 + b^2   =16

a^2 + b^2  = 9

 

a^3 + b^3 = 

(a + b) ( a^2 - ab + b^2)  = 

(a + b) ( a^2 + b^2   - ab)  =

(4)  (9 - 7/2)  = 

(4) ( 11/2) =

22

 

 

cool cool cool

 Feb 1, 2024
 #5
avatar+77 
0

Thank you so much, CPhill!!!!

NotLatePY  Feb 3, 2024
 #6
avatar+129895 
+2

4. 

 

( 1x + 4) (1x + 3)  = 1x^2  + 7x + 12

 

 

cool cool cool

 Feb 3, 2024
 #7
avatar+77 
0

Thanks a lot for the answer, but I think using only one of each number is allowed... Thanks again though!

NotLatePY  Feb 5, 2024
 #8
avatar+129895 
+1

OK....how ' bout  .....

 

(3x + - 5) (4x + 7)  =  12x^2  + 1x + - 35 

 

 

cool cool cool 

 Feb 5, 2024
 #9
avatar+129895 
+1

2. I assume the quadratic referred to is that in (1)

 

We will have the form   x^2  +  mx + n

 

If so

 

a^2 = 25

b^2 = 9

 

Sum of the roots  =  - m  =   (a^2 + b^2 )   =  34

Product of the roots =  a^2 * b^2     = (25 * 9)  =   225

 

The quadratic is

 

x^2  - 34x  + 225

 

cool cool cool

 Feb 5, 2024
 #10
avatar+77 
0

Wow, thank you so much! The application I'm doing this on says #2 is wrong ("You can use Vieta's formulas in reverse"), but I think I can figure it out by myself (somehow). Again, thanks a lot for spending all that time (lol)!

NotLatePY  Feb 6, 2024

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