Question 32

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waecmaths question: 

Find the values of k in the equation $6{{k}^{2}}=5k+6$

Option A: 

$\{-\tfrac{2}{3},-\tfrac{3}{2}\}$

Option B: 

$\{-\tfrac{2}{3},\tfrac{3}{2}\}$

Option C: 

$\{\tfrac{2}{3},-\tfrac{3}{2}\}$

Option D: 

$\{\tfrac{2}{3},\tfrac{3}{2}\}$

waecmaths solution: 

$\begin{align}  & 6{{k}^{2}}=5k+6 \\ & 6{{k}^{2}}-5k-6=0 \\ & 6{{k}^{2}}-9k+4k-6=0 \\ & 3k(2k-3)+2(2k-3)=0 \\ & (3k+2)(2k-3)=0 \\ & k=-\frac{2}{3}\text{ or }\frac{3}{2} \\\end{align}$

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