
Solve the following quadratic equation using factorization method:
\[7{y^2} - 32y + 16 = 0\]
Answer
499.2k+ views
Hint: For solving any quadratic equation of the form $a{x^2} + bx + c = 0$ using the factorization method, we need to split the term $bx$ in ${b_1}x + {b_2}x$ such that the sum of ${b_1}$ and ${b_2}$ is $b$ and the product of ${b_1}$ and ${b_2}$ is $a \times c$.
Then by further simplifying, the quadratic equation $a{x^2} + bx + c = 0$ can be written in the form of $(x - \alpha )(x - \beta ) = 0$.
In this equation the factors of the quadratic equation are$(x - \alpha )$ and $(x - \beta )$.
So, to determine the roots of the equation, these factors of the equation will be equal to $0$.
That is,
$(x - \alpha ) = 0$, $(x - \beta ) = 0$
Which gives,
$x = \alpha $, $x = \beta $
So, $\alpha $ and $\beta $ are the roots of the quadratic equation.
Complete step-by-step solution:
The given quadratic equation is \[7{y^2} - 32y + 16 = 0\].
Compare the given equation with the standard form of equation $a{x^2} + bx + c = 0$.
We get, $a = 7,b = - 32,c = 16$
Now, we need to split the term $by$ in ${b_1}y + {b_2}y$ such that the sum of ${b_1}$ and ${b_2}$ is $b$ and the product of ${b_1}$ and ${b_2}$ is $a \times c$.
Using the given terms, we get
\[{b_1} + {b_2} = - 32\]
\[{b_1} \times {b_2} = 7 \times 16 = 112\]
Let us take \[{b_1} = - 4\] and \[{b_2} = - 28\], since it satisfies both the above conditions,
\[{b_1} + {b_2} = ( - 4) + ( - 28) = - 32\]
\[\Rightarrow {b_1} \times {b_2} = ( - 4) \times ( - 28) = 112\]
We split $ - 32y$ term into $ - 4y$ and $ - 28y$,
\[7{y^2} - 4y - 28y + 16 = 0\]
Take $y$ common from the terms $7{y^2} - 4y$ and take $ - 4$ common from the terms $ - 28y + 16$, to obtain
\[y(7y - 4) - 4(7y - 4) = 0\]
Take the term $(7y - 4)$ common from the above equation, we get
$\Rightarrow (7y - 4)(y - 4) = 0$
The above equation can further be written as,
$(7y - 4) = 0$ or $(y - 4) = 0$
$\Rightarrow 7y = 4$ or $y = 4$
$\Rightarrow y = \dfrac{4}{7}$ or $y = 4$
So, the roots of the quadratic equation \[7{y^2} - 32y + 16 = 0\] are $\dfrac{4}{7}$ and $4$.
Note: The quadratic equation $a{x^2} + bx + c = 0$ can be factored in the form of $(x - \alpha )(x - \beta ) = 0$ where $(x - \alpha )$ and $(x - \beta )$ are the factors of the quadratic equation so $\alpha $ and $\beta $ are the roots of the quadratic equation and the values of $a$ cannot be $0$.
If the value of $a$ is $0$, then the quadratic equation will become a binomial equation.
Then by further simplifying, the quadratic equation $a{x^2} + bx + c = 0$ can be written in the form of $(x - \alpha )(x - \beta ) = 0$.
In this equation the factors of the quadratic equation are$(x - \alpha )$ and $(x - \beta )$.
So, to determine the roots of the equation, these factors of the equation will be equal to $0$.
That is,
$(x - \alpha ) = 0$, $(x - \beta ) = 0$
Which gives,
$x = \alpha $, $x = \beta $
So, $\alpha $ and $\beta $ are the roots of the quadratic equation.
Complete step-by-step solution:
The given quadratic equation is \[7{y^2} - 32y + 16 = 0\].
Compare the given equation with the standard form of equation $a{x^2} + bx + c = 0$.
We get, $a = 7,b = - 32,c = 16$
Now, we need to split the term $by$ in ${b_1}y + {b_2}y$ such that the sum of ${b_1}$ and ${b_2}$ is $b$ and the product of ${b_1}$ and ${b_2}$ is $a \times c$.
Using the given terms, we get
\[{b_1} + {b_2} = - 32\]
\[{b_1} \times {b_2} = 7 \times 16 = 112\]
Let us take \[{b_1} = - 4\] and \[{b_2} = - 28\], since it satisfies both the above conditions,
\[{b_1} + {b_2} = ( - 4) + ( - 28) = - 32\]
\[\Rightarrow {b_1} \times {b_2} = ( - 4) \times ( - 28) = 112\]
We split $ - 32y$ term into $ - 4y$ and $ - 28y$,
\[7{y^2} - 4y - 28y + 16 = 0\]
Take $y$ common from the terms $7{y^2} - 4y$ and take $ - 4$ common from the terms $ - 28y + 16$, to obtain
\[y(7y - 4) - 4(7y - 4) = 0\]
Take the term $(7y - 4)$ common from the above equation, we get
$\Rightarrow (7y - 4)(y - 4) = 0$
The above equation can further be written as,
$(7y - 4) = 0$ or $(y - 4) = 0$
$\Rightarrow 7y = 4$ or $y = 4$
$\Rightarrow y = \dfrac{4}{7}$ or $y = 4$
So, the roots of the quadratic equation \[7{y^2} - 32y + 16 = 0\] are $\dfrac{4}{7}$ and $4$.
Note: The quadratic equation $a{x^2} + bx + c = 0$ can be factored in the form of $(x - \alpha )(x - \beta ) = 0$ where $(x - \alpha )$ and $(x - \beta )$ are the factors of the quadratic equation so $\alpha $ and $\beta $ are the roots of the quadratic equation and the values of $a$ cannot be $0$.
If the value of $a$ is $0$, then the quadratic equation will become a binomial equation.
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

10 examples of friction in our daily life

