
How do you simplify \[3{y^2} - 2y?\]
Answer
542.7k+ views
Hint:This question describes the operation of addition/ subtraction/ multiplication/ division. The final answer would be a simplified form of a given question. This question can be solved by using a quadratic equation and we can replace the term \[x\] with \[y\]. This question can also be solved by finding the greatest common factor.
Complete step by step solution:
The given problem is shown below,
\[3{y^2} - 2y?\]
This equation can also be written as,
\[3{y^2} - 2y = 0 \to \left( 1 \right)\]
We know that,
The basic form of a quadratic equation is, \[a{x^2} + bx + c = 0 \to \left( 2 \right)\](Here, \[x\] is replaced with\[y\])
So, we get
\[y = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \to \left( 3 \right)\]
By comparing the equation\[\left( 1 \right)\]and\[\left( 2 \right)\], we get
\[\left( 1 \right) \to 3{y^2} - 2y = 0\]
\[\left( 2 \right) \to a{x^2} + bx + c = 0\]
So, we get
\[a = 3,b = - 2\]and\[c = 0\].
So, the equation\[\left( 3 \right)\]becomes
\[\left( 3 \right) \to y = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
\[
y = \dfrac{{ - \left( { - 2} \right) \pm \sqrt {{{\left( { - 2} \right)}^2} - 4 \times 3 \times 0} }}{{2
\times 3}} \\
y = \dfrac{{2 \pm \sqrt 4 }}{6} \\
\]
So, we get
\[
y = \dfrac{{2 \pm 2}}{6} \\
y = \dfrac{{2\left( { - 1 \pm 1} \right)}}{6} \\
\]
\[y = \dfrac{{1 \pm 1}}{3}\]
Case: 1
\[
y = \dfrac{{1 + 1}}{3} \\
y = \dfrac{2}{3} \\
\]
Case: 2
\[
y = \dfrac{{1 - 1}}{3} \\
y = 0 \\
\]
By using the values of \[y\], we can write the following equation
\[
3y - 2 = 0 \\
y = 0 \\
\]
The above equations can also be written as,
\[\left( {3y - 2} \right).y = 0\]
When the question is \[3{y^2} - 2y = 0\], the answer becomes \[\left( {3y - 2} \right).y = 0\]. So when the question is \[3{y^2} - 2y\], the answer becomes \[\left( {3y - 2} \right).y\].
So, the final answer is
\[3{y^2} - 2y = \left( {3y - 2} \right).y\]
Note: This type of question involves the operation of addition/ subtraction/ multiplication/ division. This question can also be solved by finding the greatest common factor. In the given question we have \[3{y^2} - 2y\], the greatest common factor of the given equation is \[y\]. So, we can take \[y\] it as a common term. So, the given question can also be written as \[\left( {3y - 2} \right).y\]. By this method, we can easily solve these types of questions. Note that we shouldn’t take \[1\] as the greatest common factor to solve the question. If we take \[1\] as a greatest common factor we won’t get a simplified form of the given equation.
Complete step by step solution:
The given problem is shown below,
\[3{y^2} - 2y?\]
This equation can also be written as,
\[3{y^2} - 2y = 0 \to \left( 1 \right)\]
We know that,
The basic form of a quadratic equation is, \[a{x^2} + bx + c = 0 \to \left( 2 \right)\](Here, \[x\] is replaced with\[y\])
So, we get
\[y = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \to \left( 3 \right)\]
By comparing the equation\[\left( 1 \right)\]and\[\left( 2 \right)\], we get
\[\left( 1 \right) \to 3{y^2} - 2y = 0\]
\[\left( 2 \right) \to a{x^2} + bx + c = 0\]
So, we get
\[a = 3,b = - 2\]and\[c = 0\].
So, the equation\[\left( 3 \right)\]becomes
\[\left( 3 \right) \to y = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
\[
y = \dfrac{{ - \left( { - 2} \right) \pm \sqrt {{{\left( { - 2} \right)}^2} - 4 \times 3 \times 0} }}{{2
\times 3}} \\
y = \dfrac{{2 \pm \sqrt 4 }}{6} \\
\]
So, we get
\[
y = \dfrac{{2 \pm 2}}{6} \\
y = \dfrac{{2\left( { - 1 \pm 1} \right)}}{6} \\
\]
\[y = \dfrac{{1 \pm 1}}{3}\]
Case: 1
\[
y = \dfrac{{1 + 1}}{3} \\
y = \dfrac{2}{3} \\
\]
Case: 2
\[
y = \dfrac{{1 - 1}}{3} \\
y = 0 \\
\]
By using the values of \[y\], we can write the following equation
\[
3y - 2 = 0 \\
y = 0 \\
\]
The above equations can also be written as,
\[\left( {3y - 2} \right).y = 0\]
When the question is \[3{y^2} - 2y = 0\], the answer becomes \[\left( {3y - 2} \right).y = 0\]. So when the question is \[3{y^2} - 2y\], the answer becomes \[\left( {3y - 2} \right).y\].
So, the final answer is
\[3{y^2} - 2y = \left( {3y - 2} \right).y\]
Note: This type of question involves the operation of addition/ subtraction/ multiplication/ division. This question can also be solved by finding the greatest common factor. In the given question we have \[3{y^2} - 2y\], the greatest common factor of the given equation is \[y\]. So, we can take \[y\] it as a common term. So, the given question can also be written as \[\left( {3y - 2} \right).y\]. By this method, we can easily solve these types of questions. Note that we shouldn’t take \[1\] as the greatest common factor to solve the question. If we take \[1\] as a greatest common factor we won’t get a simplified form of the given equation.
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