Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

Solve the following quadratic equation using factorization method:
\[7{y^2} - 32y + 16 = 0\]

Answer
VerifiedVerified
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.