Alan

g(4) means f(x) = 4, so 2x - 3 = 4 or x = 7/2.

Hence g(4) = 5 - 4*7/2  or g(4) = -9

If you are familiar with matrix solutions of simultaneous linear equations, the following matrix solution is an alternative way of tackling this problem:

(NB  I've deliberately left out the detail and just sketched the approach for those who are familiar with matrix manipulations.)

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Rom has expanded \((x^2+\frac{1}{4})^4\)  but the expression in the original question is \((x^2+\frac{1}{x})^4\)

Given that the question asks for the constant term, Rom's version might make more sense; however, it is different from the original, for which Max's answer is correct.

John McEnroe's famous expression comes to mind!   

Are you seriously saying that you don't know how to divide a fraction by 2?

If you have 2x = 36/5 you can divide both sides by 2 to get x on its own.  Dividing 36/5 by 2 gives 18/5, because 2 goes into 36 18 times.

They label the equations.  So (1) means the equation labelled ...(1).

From (1)  t = 10

So from (2)  x + y = 10    ...(5)

Subtract (3) from (4)    y² - x² = 28  or  (y+x)(y-x) = 28   ...(6)

Use (5) in (6) to get    y - x = 28/10    or  y - x = 14/5   ...(7)

Add (5) and (7) to get    2y = 64/5  or   y = 32/5 

Subtract (7) from (5) to get    2x = 36/5   or  x = 18/5

Can you take it from there?

Find the lengths as follows:

Hint:

Because this is symmetric with respect to x and y, the maximum (and minimum) will occur when x = y. 

Hence find the maximum of \(2x\sqrt{1-x^2}\)   

"Let f(x) be a quartic polynomial with integer coefficients and four integer roots. Suppose the constant term of f(x) is 6.

(a) Is it possible for x=3 to be a root of f(x)?"

(a)   (x+1)²(x-2)(x-3)   constant term is 6

"(b) Is it possible for x=3 to be a double root of f(x)?"

(x-3)²(x-a)(x-b)   constant term is 9ab   Therefore not possible for this to be 6 if a and b are integers.