Question

Solve this equation please​

Answer 1

Step-by-step explanation:

Number 1

$log_{10}( {x}^{2} - 3x + 12 ) = 1 \\ log_{10}( {x}^{2} - 3x + 12 ) = log_{10}(10) \\ cancelling \: both \: logs \: we \: have \\ {x}^{2} - 3x + 12 = 10 \\ {x}^{2} - 3x + 12 - 10 = 0 \\ {x}^{2} - 3x + 2 = 0$

$solve \: \: by \: factorisation \: method \\ {x}^{2} - 2x - x + 2 \\ {x}(x - 2) \: - 1(x - 2) \\ (x - 2)(x - 1) \\ therefore \: x = 2 \: or \: x = 1$

Number 2