CPHILL PLEASE HELP

Question

CPHILL PLEASE HELP

+1

39

10

+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

NotLatePY Feb 1, 2024

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CPhill · Answer 1 · 2024-02-01T02:23:36

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

CPhill · Answer 2 · 2024-02-03T10:05:58

#6

+130477

+2

4.

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

CPhill Feb 3, 2024

#7

+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

CPhill · Answer 3 · 2024-02-05T10:23:40

OK....how ' bout .....

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

CPhill · Answer 4 · 2024-02-05T10:32:11

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

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