Question

Solve for x, 4^x- 3×2^(x+2)+2^5=0

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Answer 1

Solve the equation x3 − 3x2 – 2x + 4 = 0

We put the numbers that are factors of 4 into the equation to see if any of them are correct.

    f(1) = 13 − 3×12 – 2×1 + 4 = 0                        1 is a solution

    f(−1) = (−1)3 − 3×(−1)2 – 2×(−1) + 4 = 2     

    f(2) = 23 − 3×22 – 2×2 + 4 = −4

    f(−2) = (−2)3 − 3×(−2)2 – 2×(−2) + 4 = −12

    f(4) = 43 − 3×42 – 2×4 + 4 = 12

    f(−4) = (−4)3 − 3×(−4)2 – 2×(−4) + 4 = −100

The only integer solution is x = 1. When we have found one solution we don’t really need to test any other numbers because we can now solve the equation by dividing by (x − 1) and trying to solve the quadratic we get from the division.

        

Now we can factorise our expression as follows:  

    x3 − 3x2 – 2x + 4 = (x − 1)(x2 − 2x − 4) = 0

It now remains for us to solve the quadratic equation.  

    x2 − 2x − 4 = 0

We use the formula for quadratics with a = 1, b = −2 and c = −4.



We have now found all three solutions of the equation x3 − 3x2 – 2x + 4 = 0. They are: eftirfarandi:

    x = 1

    x = 1 + Ö5

    x = 1 − Ö5