If $a - b = 7$ and $a b = 18$ ; find $a + b$

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Solution

The correct option is B $\pm 11$
Given,
$a - b = 7, a b = 108$
Therefore,
$(a + b)^{2} - (a - b)^{2} = 4 a b$
$=> (a + b)^{2} - 7^{2} - = 4 \times 18$
$=> (a + b)^{2} - 49 - = 72$
$=> (a + b)^{2} = 72 + 49$
$=> (a + b)^{2} = 121$
$=> a + b = \sqrt{1} 21$
$=> a + b = \pm 11$

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Similar questions

Q. The average frequency is

18

. Find the missing frequency

f

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$13 - 15$	$6$
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$21 - 23$	$5$
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The following distribution shows the daily pocket allowance of children of a locality. If the mean pocket allowance is Rs. 18, find the missing frequency f.
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If the mean of the data given below is 18, find R ?

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If a−b=7 and ab=18; find a+b

The correct option is B ±11Given,a−b=7,ab=108Therefore,(a+b)2−(a−b)2=4ab=>(a+b)2−72−=4×18=>(a+b)2−49−=72=>(a+b)2=72+49=>(a+b)2=121=>a+b=√121=>a+b=±11

If $a - b = 7$ and $a b = 18$ ; find $a + b$

The correct option is B $\pm 11$
Given,
$a - b = 7, a b = 108$
Therefore,
$(a + b)^{2} - (a - b)^{2} = 4 a b$
$=> (a + b)^{2} - 7^{2} - = 4 \times 18$
$=> (a + b)^{2} - 49 - = 72$
$=> (a + b)^{2} = 72 + 49$
$=> (a + b)^{2} = 121$
$=> a + b = \sqrt{1} 21$
$=> a + b = \pm 11$