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Question: The probability of choosing randomly a number \(c\) from set \(\left\\{ {1,2,3,4,.......,9} \right\\...

The probability of choosing randomly a number cc from set \left\\{ {1,2,3,4,.......,9} \right\\} such that the quadratic equation x2+4x+c{x^2} + 4x + c has real roots
a)19 b)29 c)39 d)49  a)\,\dfrac{1}{9} \\\ b)\,\dfrac{2}{9} \\\ c)\,\dfrac{3}{9} \\\ d)\,\dfrac{4}{9} \\\

Explanation

Solution

For any given quadratic equation ax2+bx+c=0a{x^2} + bx + c = 0 the condition for real roots is given by D0D \geqslant 0 where D=b24acD = {b^2} - 4ac. Using this concept we try to find the possible values of c and finally find the probability which is our required answer..

Complete step-by-step answer:
Question is saying that there is a number cc which belong to set \left\\{ {1,2,3,4,.......,9} \right\\} and satisfy the quadratic equation x2+4x+c=0{x^2} + 4x + c = 0 such that it has real roots
So first of all we would let the quadratic equation as we know for a given quadratic equation ax2+bx+c=0a{x^2} + bx + c = 0 has real roots then it must satisfy D0D \geqslant 0 where D=b24acD = {b^2} - 4ac.
So here the quadratic equation is x2+4x+c=0{x^2} + 4x + c = 0
Comparing with general quadratic equation ax2+bx+c=0ax^2+bx+c=0
We get a=1a=1,b=4b=4,c=1c=1
Now, D0D \geqslant 0 . So
b24ac0 (4)24(1)c0 164c0 164c c4  {b^2} - 4ac \geqslant 0 \\\ {(4)^2} - 4(1)c \geqslant 0 \\\ 16 - 4c \geqslant 0 \\\ 16 \geqslant 4c \\\ c \leqslant 4 \\\
So from the quadratic equation we came to know that cc must be less than or equal to 44
Now solving the above part that is of set cc also belongs to the set which contains \left\\{ {1,2,3,4,.......,9} \right\\} and also we know that cc must be less than or equal to 44
So here cc must be cc \left\\{ {1,2,3,4} \right\\} it means total number of favourable cases is 44
Now we are asked to find the probability of choosing cc from set \left\\{ {1,2,3,4,.......,9} \right\\}
So probability of choosing c=No.offavourablecasesTotalno.ofcasesc = \dfrac{{No.\,of\,favourable\,cases}}{{Total\,no.\,of\,cases}}
Here total no. of cases is the total number of elements in the set that is 99
So probability is 49\dfrac{4}{9}

So, the correct answer is “Option D”.

Note: For finding the probability we first need to find the total number of favourable cases and then divide it by total number of cases. Here the favourable cases are 4asc44\,as\,c \leqslant 4 and cc belongs to \left\\{ {1,2,3,4,.......,9} \right\\},So here cc can be 1,2,3,41,2,3,4.Students should remember the the condition for real roots for a quadratic equation ax2+bx+c=0a{x^2} + bx + c = 0 which is given by D0D \geqslant 0 where D=b24acD = {b^2} - 4ac.