
How do you factor the expression \[2{x^2} + 3x + 1 = 0\]?
Answer
558k+ views
Hint:
Here we are given a quadratic equation that is \[2{x^2} + 3x + 1 = 0\] and we are asked the method of factor or solve it that means we have to solve and get the values of \[x\] so here we need to break the mid-term the term containing related to variable \[x\] that the broken terms on adding or subtracting as per the condition depicted while solving the equation gives the midterm and when multiplied is equal to the product that we get by multiplying constant term and the \[{x^2}\] term.
Complete step by step solution:
Here we are given an equation that is \[2{x^2} + 3x + 1 = 0\]and we are asked to factor it.
So above given equation is the quadratic equation (a quadratic equation is the equation in which the highest of any variable is\[2\])
So for solving it we need to basically find the values of \[x\]that satisfies the equation completely for which we need to breakdown the mid-term (the term containing related to variable \[x\])in such a way that the broken terms are just the representation of midterm and addition or subtraction gives the midterm and when multiplied is equal to the product that we get by multiplying constant term and the \[{x^2}\]term
Now the given equation is –
\[2{x^2} + 3x + 1 = 0\]
Breaking down the midterm we get-
\[2{x^2} + 2x + 1x + 1 = 0\]-
On further simplifying
\[2x(1x + 1) + 1(1x + 1)\]
\[\left( {2x + 1} \right)(x + 1) = 0\]
Now comparing each one of with zero that comes out to be –
\[
\left( {2x + 1} \right)(x + 1) = 0 \\
\left( {x + 1} \right) = 0 \\
x = - 1 \\
and \\
(2x + 1) = 0 \\
x = \dfrac{{ - 1}}{2} \\
\]
Now the the values of \[x\] are \[x = \dfrac{{ - 1}}{2}, - 1\] that are the factors for the given equation that is \[2{x^2} + 3x + 1 = 0\]
Note:
While solving such kind of questions the splitting of the midterm should be done correctly because it is driving step of the solution to this question also if the midterm cannot be splitted try other methods like completing the square method in which u need to eliminate the coefficient of the term containing the \[{x^2}\] making the coefficient related to the \[{x^2}\] equals \[1\] then looking the coefficient of the term containing \[x\] making its coefficient into half and squaring the coefficient then adding and subtracting the whole equation by that number hence further solving gives us the roots of the quadratic equation .
Here we are given a quadratic equation that is \[2{x^2} + 3x + 1 = 0\] and we are asked the method of factor or solve it that means we have to solve and get the values of \[x\] so here we need to break the mid-term the term containing related to variable \[x\] that the broken terms on adding or subtracting as per the condition depicted while solving the equation gives the midterm and when multiplied is equal to the product that we get by multiplying constant term and the \[{x^2}\] term.
Complete step by step solution:
Here we are given an equation that is \[2{x^2} + 3x + 1 = 0\]and we are asked to factor it.
So above given equation is the quadratic equation (a quadratic equation is the equation in which the highest of any variable is\[2\])
So for solving it we need to basically find the values of \[x\]that satisfies the equation completely for which we need to breakdown the mid-term (the term containing related to variable \[x\])in such a way that the broken terms are just the representation of midterm and addition or subtraction gives the midterm and when multiplied is equal to the product that we get by multiplying constant term and the \[{x^2}\]term
Now the given equation is –
\[2{x^2} + 3x + 1 = 0\]
Breaking down the midterm we get-
\[2{x^2} + 2x + 1x + 1 = 0\]-
On further simplifying
\[2x(1x + 1) + 1(1x + 1)\]
\[\left( {2x + 1} \right)(x + 1) = 0\]
Now comparing each one of with zero that comes out to be –
\[
\left( {2x + 1} \right)(x + 1) = 0 \\
\left( {x + 1} \right) = 0 \\
x = - 1 \\
and \\
(2x + 1) = 0 \\
x = \dfrac{{ - 1}}{2} \\
\]
Now the the values of \[x\] are \[x = \dfrac{{ - 1}}{2}, - 1\] that are the factors for the given equation that is \[2{x^2} + 3x + 1 = 0\]
Note:
While solving such kind of questions the splitting of the midterm should be done correctly because it is driving step of the solution to this question also if the midterm cannot be splitted try other methods like completing the square method in which u need to eliminate the coefficient of the term containing the \[{x^2}\] making the coefficient related to the \[{x^2}\] equals \[1\] then looking the coefficient of the term containing \[x\] making its coefficient into half and squaring the coefficient then adding and subtracting the whole equation by that number hence further solving gives us the roots of the quadratic equation .
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