
Roots of the quadratic equation $2{x^2} - 6x = 0$ are
A - 0,3
B - 0, - 3
C - 2,6
D - 0,2
Answer
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Hint: This sum is from the chapter Quadratic Equations. In this one thing the student should keep in mind is while taking out common variables and cancelling them off. Since this is a linear equation with degree $2$, there has to be $2$ roots. But if the student strikes off the common variable i.e. $x$ in this particular sum , he would get only one root. But the student should keep in mind that the root which he cancelled is actually $0$. Another way to solve this particular sum is by taking the sum of the roots and product of the roots and then compare the values. This is the preferred way to solve any quadratic equation as it makes the approach easy and clear.
Complete answer:
Standard quadratic equation with degree $2$ is of the form $a{x^2} + bx + c$.
Sum of the roots for quadratic equation with degree $2$ is given by $\dfrac{{ - b}}{a}$
Sum of the roots for quadratic equation with degree $2$ is given by $\dfrac{c}{a}$
Bringing the given problem in the standard form of quadratic equation
$\Rightarrow$ $2{x^2} - 6x - 0$
Comparing the given numerical with the standard form of quadratic equation we get,
$\Rightarrow$ Value of $a = 2$
$\Rightarrow$ Value of $b = 6$
$\Rightarrow$ Value of $c = 0$
Let ${a_1}\ & {a_2}$ be roots of the quadratic equation $2{x^2} - 6x - 0$.
$\therefore $ Sum of the Roots $ = {a_1} + {a_2}$
$\therefore {a_1} + {a_2} = \dfrac{{ - ( - 6)}}{2} = 3..........(1)$
Product of the Roots $ = {a_1}{a_2}$
$\Rightarrow$ ${a_1}{a_2} = \dfrac{0}{2}..........(2)$
From Equation $1\& 2$ we can say that the roots for the quadratic equation are $0,3$.
Thus the answer for the given question is $A - 0,3$
Note: It is always preferable to solve the sum with the above method . Students can solve the quadratic equation of degree more than $2$ as well with this method. This makes the sum easy to understand and there is very little chance of going wrong. Thus students are advised not to cancel off the common terms when it comes to solving quadratic equations.
Complete answer:
Standard quadratic equation with degree $2$ is of the form $a{x^2} + bx + c$.
Sum of the roots for quadratic equation with degree $2$ is given by $\dfrac{{ - b}}{a}$
Sum of the roots for quadratic equation with degree $2$ is given by $\dfrac{c}{a}$
Bringing the given problem in the standard form of quadratic equation
$\Rightarrow$ $2{x^2} - 6x - 0$
Comparing the given numerical with the standard form of quadratic equation we get,
$\Rightarrow$ Value of $a = 2$
$\Rightarrow$ Value of $b = 6$
$\Rightarrow$ Value of $c = 0$
Let ${a_1}\ & {a_2}$ be roots of the quadratic equation $2{x^2} - 6x - 0$.
$\therefore $ Sum of the Roots $ = {a_1} + {a_2}$
$\therefore {a_1} + {a_2} = \dfrac{{ - ( - 6)}}{2} = 3..........(1)$
Product of the Roots $ = {a_1}{a_2}$
$\Rightarrow$ ${a_1}{a_2} = \dfrac{0}{2}..........(2)$
From Equation $1\& 2$ we can say that the roots for the quadratic equation are $0,3$.
Thus the answer for the given question is $A - 0,3$
Note: It is always preferable to solve the sum with the above method . Students can solve the quadratic equation of degree more than $2$ as well with this method. This makes the sum easy to understand and there is very little chance of going wrong. Thus students are advised not to cancel off the common terms when it comes to solving quadratic equations.
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