How do you solve \[{y^2} - 6y + 8 = 0\]?
Answer
574.8k+ views
Hint: In these type polynomials, we will solve by using splitting middle term, for a polynomial of the form \[a{x^2} + bx + c = 0\], rewrite the middle term as a sum of two terms whose product is \[a \cdot c = 1 \cdot 8 = 8\] and whose sum is \[b = - 6\], when you solve the expression we will get the required values.
Complete step-by-step answer:
Now given equation is \[{y^2} - 6y + 8 = 0\],
This can be factorised by splitting the middle term, now rewrite the middle term as a sum of two terms, we will get two terms. Now sum of two terms whose product is \[a \cdot c = 1 \cdot 8 = 8\] and whose sum is \[b = - 6\]
Now we will find the numbers which satisfy the given data, i.e., sum of the numbers should be equal to -6 and their product should be equal to 8,
So, now rewrite the -6 as -4 and -2, whose product will be equal to 8 and their sum is -6.
Now using distributive property, we get
\[ \Rightarrow {y^2} - 4y - 2y + 8 = 0\],
By grouping the first two terms and last two terms, we get,
\[ \Rightarrow \left( {{y^2} - 4y} \right) + \left( { - 2y + 8} \right) = 0\],
Now factor out the highest common factor, we get
\[ \Rightarrow y\left( {y - 4} \right) - 2\left( {y - 4} \right) = 0\],
Now taking common term in both, we get,
\[ \Rightarrow \left( {y - 4} \right)\left( {y - 2} \right) = 0\]
So, the equating terms to zero, so that we will get the values of \[y\],i.e.,
\[ \Rightarrow \]\[y - 4 = 0\] and \[y - 2 = 0\],
So simplifying we get,
\[ \Rightarrow y = 2\] and \[y = 4\],
So the values of \[y\] are 2 and 4.
\[\therefore \]The values for \[y\] when the polynomial \[{y^2} - 6y + 8 = 0\] is solved are 2 and 4.
Note:
We have several options for factoring when you are solving the polynomial equations. In a polynomial, irrespective of how many terms it has, we should always check the highest common factors first. The highest common factor is our biggest expression which would go into all our terms, and when we use H.C.F it is similar to performing the distributive property backwards. If the expression is a binomial then we must look for the differences of squares, difference of cubes, or even the sum of cubes, finally once the polynomial is factored fully, you can then use the zero property for solving the equation.
Complete step-by-step answer:
Now given equation is \[{y^2} - 6y + 8 = 0\],
This can be factorised by splitting the middle term, now rewrite the middle term as a sum of two terms, we will get two terms. Now sum of two terms whose product is \[a \cdot c = 1 \cdot 8 = 8\] and whose sum is \[b = - 6\]
Now we will find the numbers which satisfy the given data, i.e., sum of the numbers should be equal to -6 and their product should be equal to 8,
So, now rewrite the -6 as -4 and -2, whose product will be equal to 8 and their sum is -6.
Now using distributive property, we get
\[ \Rightarrow {y^2} - 4y - 2y + 8 = 0\],
By grouping the first two terms and last two terms, we get,
\[ \Rightarrow \left( {{y^2} - 4y} \right) + \left( { - 2y + 8} \right) = 0\],
Now factor out the highest common factor, we get
\[ \Rightarrow y\left( {y - 4} \right) - 2\left( {y - 4} \right) = 0\],
Now taking common term in both, we get,
\[ \Rightarrow \left( {y - 4} \right)\left( {y - 2} \right) = 0\]
So, the equating terms to zero, so that we will get the values of \[y\],i.e.,
\[ \Rightarrow \]\[y - 4 = 0\] and \[y - 2 = 0\],
So simplifying we get,
\[ \Rightarrow y = 2\] and \[y = 4\],
So the values of \[y\] are 2 and 4.
\[\therefore \]The values for \[y\] when the polynomial \[{y^2} - 6y + 8 = 0\] is solved are 2 and 4.
Note:
We have several options for factoring when you are solving the polynomial equations. In a polynomial, irrespective of how many terms it has, we should always check the highest common factors first. The highest common factor is our biggest expression which would go into all our terms, and when we use H.C.F it is similar to performing the distributive property backwards. If the expression is a binomial then we must look for the differences of squares, difference of cubes, or even the sum of cubes, finally once the polynomial is factored fully, you can then use the zero property for solving the equation.
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

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

Master Class 11 Chemistry: Engaging Questions & Answers for Success

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

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Which among the following are examples of coming together class 11 social science CBSE

