We know that in a quadratic equation ax^2 + bx + c = 0
Product of roots = c/a
Here we have the equation 5x^2 + 13x + k = 0
Product of roots = k/5
Given that the roots are reciprocals of each other. So if one root is p, the other would be 1/p. So, their product will always be 1.
=> 1 = k/5
=> k = 5.
Hence, the value of k is 5.
Product of roots = c/a
Here we have the equation 5x^2 + 13x + k = 0
Product of roots = k/5
Given that the roots are reciprocals of each other. So if one root is p, the other would be 1/p. So, their product will always be 1.
=> 1 = k/5
=> k = 5.
Hence, the value of k is 5.
nice one ,thanks
ReplyDeleteThe no. Of words made by rearranging the letters of the word apurba so that vowels and consonants. Are alternate is??
ReplyDelete'Apurba' has letters a, a, b, p, r, u
ReplyDeleteThe vowels we have are a, a, u.
b, p, r are consonants.
We have to find the number of words using these letters such that vowels are consonants take alternate places.
Think of each letter's place as an empty space that we'll fill with letters.
[ ][ ][ ][ ][ ][ ]
Now, we'll take the first case, in which the first letter of the word is a vowel.
CASE 1 - FIRST LETTER IS A VOWEL
we have 3 vowels a, a, u available to fill in space 1. So the number of ways in which space 1 can be filled is 3
[ 3 ][ ][ ][ ][ ][ ]
Since the first space has been taken by a consonant, the next place must be occupied by a consonant, as the question demands. We have three consonants b, p, r to fill space 2. So the number of ways in which space 2 can be filled is 3.
[ 3 ][ 3 ][ ][ ][ ][ ]
Space 3 demands a vowel, but since one vowel has already been put in space 1, we have only two possible ways to fill space 3.
[ 3 ][ 3 ][ 2 ][ ][ ][ ]
Similarly, space 4 asks for a consonant and as we have only 2 of them left, the number if ways to fill space 4 becomes 2.
[ 3 ][ 2 ][ 2 ][ 2 ][ ][ ]
We need a vowel again for space 5, and there only 1 left now. So there's only 1 way to fill space 5.
[ 3 ][ 3 ][ 2 ][ 2 ][ 1 ][ ]
Similarly, space 6 seeks a consonant, and there's just one last one left. Just 1 way to fill space 6.
We have the number of ways letters can be put in their places to fulfill the conditions as [ 3 ][ 3 ][ 2 ][ 2 ][ 1 ][ 1 ]. Hence, the number of words that can be formed satisfying the given conditions become 3 x 3 x 2 x 2 x 1 x 1 = 36 (This is simple permutations and combinations)
now, well take case 2
CASE 2 - FIRST LETTER IS A CONSONANT
Retrace what you did in case 1. The only difference you'll find is that the first pace is taken by a consonant and the next by a vowel and the same route follows. In the end, you'll the number of wrote possible for this case to be the same i.e. 3 x 3 x 2 x 2 x 1 x 1 = 36.
Hence the total number of words possible as per the conditions given in the question will be 36 + 36 =72.
The first three of four given numbers are in g.p. And. Their last three are. In a.p. With common differennce 6.if. First and. Fourth numbers are equal then first. Number is?
ReplyDeleteit was helpful..thanx
ReplyDeleteThe first and last numbers are given to be equal. Let's assume them to be a.
ReplyDeleteThe last three numbers are said to be in A.P. and the common difference given is 6. So, the second and third numbers will be (a - 12) and (a - 6) respectively.
Now, since it is given that the first three numbers are in G.P. (2nd no/1st no) should be equal to (3rd no/2nd no).
=> (a - 12)/a = (a - 6)/(a - 12)
=> (a - 12)^2 = a(a - 6)
=> a^2 - 24a + 144 = a^2 - 6a
=> - 24a + 6a = -144
=> -18a = -144
=> a = 8
The rest of the numbers can be easily determined now. The four numbers thus come out to be 8, -4, 2, 8.
Very. True. Sir
ReplyDeleteA very big thanks sir...........
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Please don't be. Just comment what part of the problem/solution you had trouble with so that I can help you better. :)
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ReplyDelete12
ReplyDeleteWhy k/5 =1? If 1 root is p other is 1 upon p product = 1 how
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