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Sonja [21]
3 years ago
8

What are the solution(s) to the quadratic equation x2 - 25 = 0?

Mathematics
1 answer:
FromTheMoon [43]3 years ago
7 0

Answer:

A

Step-by-step explanation:

Given

x² - 25 = 0 ( add 25 to both sides )

x² = 25 ( take the square root of both sides )

x = ± \sqrt{25} = ± 5

Solutions are x = 5 and x = - 5

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For some constants a and b let \[f(x) = \left\{ \begin{array}{cl} 9 - 2x &amp; \text{if } x \le 3, \\ ax + b &amp; \text{if } x
Viktor [21]

Answer:

The value of a+b is 4.

Step-by-step explanation:

The given function is

\[f(x) = \left\{ \begin{array}{cl} 9 - 2x & \text{if } x \le 3, \\ ax + b & \text{if } x > 3. \end{array} \right.\]

It is given that for some constants a and b the function f has the property that f(f(x))=x for all x.

For x≤3,

f(x)=9-2x

For x>3,

f(x)=ax+b

At x=0,

f(0)=9-2(0)=9

f(f(0))=f(9)\Rightarrow a(9)+b=9a+b

Using property f(f(x))=x,

f(f(0))=0

9a+b=0                     .... (1)

At x=1,

f(1)=9-2(1)=7

f(f(1))=f(7)\Rightarrow a(7)+b=7a+b

Using property f(f(x))=x,

f(f(1))=1

7a+b=1                     .... (2)

Subtract equation (2) from equation (1).

9a+b-(7a+b)=0-1    

2a=-1

Divide both sides by 2.

a=-\frac{1}{2}

Substitute this value in equation (1).

9(-\frac{1}{2})+b=0

b=\frac{9}{2}

The value of a is -\frac{1}{2} and value of b is \frac{9}{2}. The value of a+b is

a+b=-\frac{1}{2}+\frac{9}{2}

a+b=4

Therefore the value of a+b is 4.

5 0
3 years ago
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