HOW DO YOU SOLVE 32x − 10 × 3x + 9 = 0?
Substitute u = 3^x. The equation becomes u² − 10u + 9 = 0, which factorises to (u − 1)(u − 9) = 0, giving 3^x = 1 and 3^x = 9, so x = 0 and x = 2 and the two solutions sum to 2.
The equation 32x − 10 × 3x + 9 = 0 has two real solutions. What is the sum of those solutions?
The first term is the square of the second, which is the signature of a hidden quadratic. Substitute, solve in the new variable, and only then convert back.
- 32x = (3x)2, so put u = 3x
- u2 − 10u + 9 = 0
- (u − 1)(u − 9) = 0, so u = 1 or u = 9
- 3x = 1 gives x = 0; 3x = 9 gives x = 2
- sum = 0 + 2 = 2
10 is what you get if you stop at the substitution and quote the sum of the u values instead of the x values. The sum of the roots of u² − 10u + 9 really is 10, which is exactly why the number is offered: it is the right answer to the wrong question.
9 is what you get if you read 3^(2x) as 2 × 3^x which turns the equation into a linear one in 3^x. An exponent of 2x means the whole power is squared, never doubled.
You never need the individual roots. Because x = log₃u, the sum of the x values is log₃(u₁) + log₃(u₂) = log₃(u₁u₂), and the product of the roots of the quadratic is just the constant term over the leading one, which is 9. So the answer is log₃9 = 2 straight from the coefficients, without factorising anything.
THE GENERAL RULE
Whenever an equation contains a^(2x) and a^x, or x⁴ and x², or tan²θ and tanθ, it is a quadratic wearing a costume. Substitute, solve, and then — this is the part the question is testing — remember to convert back. Every one of these is written so that the un-converted answer is one of the options.
Updated · written to the published ESAT content specification