can I get help with this short problem please. (grade 8th quastion)

[bamba12312]

prove that b = -10a
just look at the photo it is not hard to understand.
thank you .

 

[mmm4444bot]

it is not hard to understand

Hello. Very good. Please explain what you do understand so far. Where are you stuck? Please follow the posting guidelines.

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[Dr.Peterson]

View attachment 33939
prove that b = -10a
just look at the photo it is not hard to understand.
thank you .

The problem is not quite clearly stated. My guess would be that what they call the "definition area" means the entire natural domain of the function (rather than, say, an artificially restricted domain).

How would you find the (natural) domain of f(x)?

 

[bamba12312]

Hello. Very good. Please explain what you do understand so far. Where are you stuck? Please follow the posting guidelines.

Posting Guidelines (Summary)

Welcome to our tutoring boards! :) This page summarizes the main points from our posting guidelines. As our name implies, we provide math help (primarily to students with homework). We do not generally post immediate answers or step-by-step solutions. We don't do your homework. We prefer to...

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I solved it because the definition area is 0 =< x =< 10 so if you put 10 instead of x it will be
a(10)^2 =a100 +b10 = 0
b10 = -a100
b = -a10

 

[Dr.Peterson]

I solved it because the definition area is 0 =< x =< 10 so if you put 10 instead of x it will be
a(10)^2 =a100 +b10 = 0
b10 = -a100
b = -a10.

We normally write numerical coefficients on the left, so this is initially confusing to read; but what you have done is correct.

I would express it like this for clarity:

Since the graph shows that f(x) = 0 at each extremity of the domain, and clearly f(0) = sqrt(a(0)^2 + b(0)) = 0, we need

f(10) = sqrt(a(10)^2 + b(10)) = 0​

Squaring both sides,

a(10)^2 + b(10) = 0​

100a + 10b = 0​

10a + b = 0​

b = -10a​

Another approach, which is what I suggested, is not to depend on the graph, but just find the domain of f(x) = sqrt(ax^2 + bx). In order to take the square root, the radicand must be non-negative, so the domain is given by

ax^2 + bx >= 0​

x(ax + b) >= 0​

This will be true when either both factors on the LHS are positive, or both are negative; following a common method for solving polynomial inequalities, you can observe that the LHS is zero when x = 0 or x = -b/a, and it will be positive between those. So the domain is

0 <= x <= -b/a​

We want -b/a = 10, so b = -10a.

This is a longer method, but doesn't assume we've been given the correct graph.

 

[Steven G]

I solved it because the definition area is 0 =< x =< 10 so if you put 10 instead of x it will be
a(10)^2 =a100 +b10 = 0
b10 = -a100
b = -a10

While it is true that a(10)^2 = a(100) or 100a, it is NOT true that a(10)^2 = a(100) +b(10). If you add 10b to the right hand side of the equal sign, then you MUST add 10b to the left hand side of the equal sign.

 

[Dr.Peterson]

While it is true that a(10)^2 = a(100) or 100a, it is NOT true that a(10)^2 = a(100) +b(10). If you add 10b to the right hand side of the equal sign, then you MUST add 10b to the left hand side of the equal sign.

No, you're being confused by the less-than-ideal notation and a typo I'd failed to comment on. They didn't add 10b to one side.

a(10)^2 was supposed to be f(10)^2. So that first line means

[f(10)]^2 = [sqrt(a*(10)^2 + b*(10))]^2 = a*100 + b*10 = 0​

 

[Steven G]

No, you're being confused by the less-than-ideal notation and a typo I'd failed to comment on. They didn't add 10b to one side.

a(10)^2 was supposed to be f(10)^2. So that first line means

[f(10)]^2 = [sqrt(a*(10)^2 + b*(10))]^2 = a*100 + b*10 = 0​

I wonder if you're right, especially since the OP failed to write sqrt after f(10^2)

Many students, as you know, write things like:
9+8+6+4
=17+6
=23+4
=27

 

[Dr.Peterson]

I wonder if you're right, especially since the OP failed to write sqrt after f(10^2)

Many students, as you know, write things like:
9+8+6+4
=17+6
=23+4
=27

You're right about the common misuse of "="; but the OP never did write f(10^2), or even a(10^2)! I don't think the mistake you refer to has been made here at all.

I solved it because the definition area is 0 =< x =< 10 so if you put 10 instead of x it will be
a(10)^2 =a100 +b10 = 0
b10 = -a100
b = -a10

This was certainly intended to be f(10)^2, which correctly means the output squared.