Unit 1/2 Questions (quadratics)

[#1procrastinator]

1) Does anyone know how to find x in 2^x = 3 + x without a CAS?

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2)

y = 0.125x^2 + 1.25x + 3
y = 11 + bx

Find b and then find the points of intersection

I got b = 3.25 and b = -0.75, but the book says only b = -0.75 is correct. Then for the POIs, only the ones you get from plugging in -0.75 are listed (if there are two answers, the book lists them both) (8, 5)

Now there's a diagram accompaniment so I know after using both b's that -0.75 is correct because 3.25 gives negative values and there are no negatives in the diagram. S

Assuming I haven't made any errors, my questions is there a way of knowing which value is correct without having to plug in and solve both? (because that question WAS separate and before the POI one) It doesn't make sense to me that you gotta put in two answers and when you do the next question, only one of them turns out to be correct

(this is from Maths Quest by the way, Ex 2L Q. 10c)

Thanks


EDIT:

Also this one:

y=x^2 + bx -14 and y = -9x + c intersect at (-1, and (-3, 10). Find b and c

For the first pair, I got b = -5 and c = -17 and for the second b = -21 and c = -1

When you plug them back in, why does only the first set give the correct x values? (assuming, once again, I haven't made any errors)

[taiga]

First one can't be done by hand, I think x is just a function of x in that case. (Don't have CAS on me to show what that gives)

What method did you use to find the points for b?

[#1procrastinator]

If I remember correctly, I used the discriminant

I'll get back to you lol

[#1procrastinator]

Sorry, I forgot the negative sign in front of the 0.125x^2 in the first question

-0.125x^2 + 1.25x + 3 = 11+bx

-> -0.125x^2 + (1.25 - b)x - 8

-> (1.25 - b)^2 - 4 * -0.125 * -8

-> 1.5625 - 2.5b + b^2 - 4

-> b^2 - 2.5b - 2.4375

...input into quadratic formula

[#1procrastinator]

Why can't the exponent/log one been done by hand?

[brightsky]

transcendental function. you can never get one x on one side and the other numbers on the other. there are ways to approximate the answer, which as far as i know is beyond the mm course and therefore would not be tested.

[taiga]

Given you probably found this in the quadratics portion of your book

2^x = 3 + x may be a typo and it is really x^2 = 3 + x in which case you can use the quadratic equation to find the answers without a calculator.

Otherwise the question is looking for you to identify that this is not a quadratic and therefore can not be solved with any analytic methods that you have learnt. This can be solved with "numerical methods" which is basically guessing the range it is in and then plugging in values to see which x value is closest to satisfying the equation, but I don't think you deal with that at all in VCE (except to a slight extent in GMA as I've heard).

[#1procrastinator]

Oh, ok. That was under logarithms, the question said to use a CAS but I was curious to see if it could be solved by hand

So for the other 2 questions, did I do something wrong?