Find the value of $k$ for which the equation $3{x^2} - 6x + k = 0$ has a distinct and real root.
Answer
628.5k+ views
Hint:
Here, we will use the property of discriminant when the roots are distinct and real. Then we will compare the given quadratic equation with the general equation and find all the coefficients. Then we will substitute the value of the coefficient in the formula of discriminant and solve it further to get the required value.
Formula Used:
We will use the formula: ${\text{D = }}{b^2} - 4ac$
Complete step by step solution:
A quadratic equation can be written in a form of $a{x^2} + bx + c = 0$
As we know, if a quadratic equation has distinct and real roots then, discriminant is greater than 0.
${\text{D}} > 0$
We know that the formula of Discriminant, ${\text{D = }}{b^2} - 4ac$. Therefore,
$ \Rightarrow {b^2} - 4ac > 0$
Now, we have to find the value of $k$ for which the equation $3{x^2} - 6x + k = 0$ has distinct and real roots.
Comparing this equation with $a{x^2} + bx + c = 0$, we get
$a = 3{\text{, }}b = - 6,c = k$
Now, using the formula ${\text{D = }}{b^2} - 4ac$, we get
$ \Rightarrow {\text{D = }}{\left( { - 6} \right)^2} - 4\left( 3 \right)\left( k \right)$
Simplifying the expression, we get
${\text{D = }}36 - 12k$
But, it is given that the roots are distinct and equal,
Thus,
${\text{D = }}36 - 12k > 0$
Adding $12k$ on both the sides of the inequality,
$ \Rightarrow 36 > 12k$
Dividing both sides by 12, we get,
$ \Rightarrow k < 3$
Hence, this is the required value of $k$
Note:
A quadratic equation is an equation which has the highest degree of variable as 2 and has only 2 solutions. While solving this question, we should know the formula and properties of Discriminant. We need to keep in mind the following conditions:
1) We get two distinct roots i.e. positive and different when ${\text{D}} > 0$.
2) We get to equal roots i.e. positive and same when, ${\text{D}} = 0$.
3) We get no real roots i.e. negative roots when, ${\text{D}} < 0$.
As here the roots were real and distinct, so we used the condition ${\text{D}} > 0$
Here, we will use the property of discriminant when the roots are distinct and real. Then we will compare the given quadratic equation with the general equation and find all the coefficients. Then we will substitute the value of the coefficient in the formula of discriminant and solve it further to get the required value.
Formula Used:
We will use the formula: ${\text{D = }}{b^2} - 4ac$
Complete step by step solution:
A quadratic equation can be written in a form of $a{x^2} + bx + c = 0$
As we know, if a quadratic equation has distinct and real roots then, discriminant is greater than 0.
${\text{D}} > 0$
We know that the formula of Discriminant, ${\text{D = }}{b^2} - 4ac$. Therefore,
$ \Rightarrow {b^2} - 4ac > 0$
Now, we have to find the value of $k$ for which the equation $3{x^2} - 6x + k = 0$ has distinct and real roots.
Comparing this equation with $a{x^2} + bx + c = 0$, we get
$a = 3{\text{, }}b = - 6,c = k$
Now, using the formula ${\text{D = }}{b^2} - 4ac$, we get
$ \Rightarrow {\text{D = }}{\left( { - 6} \right)^2} - 4\left( 3 \right)\left( k \right)$
Simplifying the expression, we get
${\text{D = }}36 - 12k$
But, it is given that the roots are distinct and equal,
Thus,
${\text{D = }}36 - 12k > 0$
Adding $12k$ on both the sides of the inequality,
$ \Rightarrow 36 > 12k$
Dividing both sides by 12, we get,
$ \Rightarrow k < 3$
Hence, this is the required value of $k$
Note:
A quadratic equation is an equation which has the highest degree of variable as 2 and has only 2 solutions. While solving this question, we should know the formula and properties of Discriminant. We need to keep in mind the following conditions:
1) We get two distinct roots i.e. positive and different when ${\text{D}} > 0$.
2) We get to equal roots i.e. positive and same when, ${\text{D}} = 0$.
3) We get no real roots i.e. negative roots when, ${\text{D}} < 0$.
As here the roots were real and distinct, so we used the condition ${\text{D}} > 0$
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