Answer
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Hint:For any given quadratic equation $a{x^2} + bx + c = 0$ the condition for real roots is given by $D \geqslant 0$ where $D = {b^2} - 4ac$. 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 $c$ which belong to set $\left\{ {1,2,3,4,.......,9} \right\}$ and satisfy the quadratic equation ${x^2} + 4x + c = 0$ such that it has real roots
So first of all we would let the quadratic equation as we know for a given quadratic equation $a{x^2} + bx + c = 0$ has real roots then it must satisfy $D \geqslant 0$ where $D = {b^2} - 4ac$.
So here the quadratic equation is ${x^2} + 4x + c = 0$
Comparing with general quadratic equation $ax^2+bx+c=0$
We get $a=1$,$b=4$,$c=1$
Now, $D \geqslant 0$ . So
$
{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 $c$ must be less than or equal to $4$
Now solving the above part that is of set $c$ also belongs to the set which contains $\left\{ {1,2,3,4,.......,9} \right\}$ and also we know that $c$ must be less than or equal to $4$
So here $c$ must be $c$ $\left\{ {1,2,3,4} \right\}$ it means total number of favourable cases is $4$
Now we are asked to find the probability of choosing $c$ from set $\left\{ {1,2,3,4,.......,9} \right\}$
So probability of choosing $c = \dfrac{{No.\,of\,favourable\,cases}}{{Total\,no.\,of\,cases}}$
Here total no. of cases is the total number of elements in the set that is $9$
So probability is $\dfrac{4}{9}$
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 $4\,as\,c \leqslant 4$ and $c$ belongs to $\left\{ {1,2,3,4,.......,9} \right\}$,So here $c$ can be $1,2,3,4$.Students should remember the the condition for real roots for a quadratic equation $a{x^2} + bx + c = 0$ which is given by $D \geqslant 0$ where $D = {b^2} - 4ac$.
Complete step-by-step answer:
Question is saying that there is a number $c$ which belong to set $\left\{ {1,2,3,4,.......,9} \right\}$ and satisfy the quadratic equation ${x^2} + 4x + c = 0$ such that it has real roots
So first of all we would let the quadratic equation as we know for a given quadratic equation $a{x^2} + bx + c = 0$ has real roots then it must satisfy $D \geqslant 0$ where $D = {b^2} - 4ac$.
So here the quadratic equation is ${x^2} + 4x + c = 0$
Comparing with general quadratic equation $ax^2+bx+c=0$
We get $a=1$,$b=4$,$c=1$
Now, $D \geqslant 0$ . So
$
{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 $c$ must be less than or equal to $4$
Now solving the above part that is of set $c$ also belongs to the set which contains $\left\{ {1,2,3,4,.......,9} \right\}$ and also we know that $c$ must be less than or equal to $4$
So here $c$ must be $c$ $\left\{ {1,2,3,4} \right\}$ it means total number of favourable cases is $4$
Now we are asked to find the probability of choosing $c$ from set $\left\{ {1,2,3,4,.......,9} \right\}$
So probability of choosing $c = \dfrac{{No.\,of\,favourable\,cases}}{{Total\,no.\,of\,cases}}$
Here total no. of cases is the total number of elements in the set that is $9$
So probability is $\dfrac{4}{9}$
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 $4\,as\,c \leqslant 4$ and $c$ belongs to $\left\{ {1,2,3,4,.......,9} \right\}$,So here $c$ can be $1,2,3,4$.Students should remember the the condition for real roots for a quadratic equation $a{x^2} + bx + c = 0$ which is given by $D \geqslant 0$ where $D = {b^2} - 4ac$.
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