Given $4 m (m + 8) = 36$ , calculate product of the two solutions to the given equation.

- 12

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- 9

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0

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9

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Solution

The correct option is A $- 9$
Given, $4 m (m + 8) = 36$
$\Rightarrow 4 m^{2} + 32 m - 36 = 0$
$\Rightarrow 4 (m^{2} + 8 m - 9 = 0)$
$\Rightarrow 4 (m^{2} + 9 m - m - 9) = 0$
$\Rightarrow 4 (m (m + 9) - 1 (m + 9)) = 0$
$\Rightarrow 4 ((m + 9) (m - 1)) = 0$
Product of two solutions $= - 9 \times 1 = - 9$

Suggest Corrections

Similar questions

Verify each of the following :

(i) $\frac{3}{7} \times \frac{- 5}{9} = \frac{- 5}{9} \times \frac{3}{7}$

(ii) $\frac{- 8}{7} \times \frac{13}{9} = \frac{13}{9} \times \frac{- 8}{7}$

(iii) $\frac{- 12}{5} \times \frac{7}{- 36} = \frac{7}{- 36} \times \frac{- 12}{5}$

(iv) $- 8 \times \frac{- 13}{12} = \frac{- 13}{12} \times (- 8)$

Units of Labour	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$
Units of Output	$0$	$4$	$12$	$24$	$32$	$36$	$36$	$28$	$16$	$2$

In the given proportion $9 : 3 :: 36 : 12$ extremes are

⎡ ⎢ ⎣ \begin{matrix} 3 & 2 & - 9 - 6 & - 4 & 18 12 & 8 & - 36 \end{matrix} ⎤ ⎥ ⎦

2 x^{2} + 2 y^{2} + 3 x + 4 y + \frac{9}{8} = 0

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