Solve this equation please

Mathematics

1 answer:

nalin [4]2 years ago

8 0

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

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If it has only one solution that means that there is only one value of x that will fulfill this equation
35 - 5x + 6x - 36 = -24 - 17x + 17x + 24
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