
Find the zeros of the quadratic polynomial $ {x^2} + 7x + 10 $ and verify the relationship between the zeros and coefficients.
Answer
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Hint: Here we will find the zeroes of the given quadratic polynomial and then will simplify the values of “x”. zeroes are also known as the roots of the polynomial. Then will find the correlation between the zeros and the coefficients by taking the sum and the product of the roots.
Complete step-by-step answer:
Take the given expression: $ {x^2} + 7x + 10 $
Split the middle term in such a way that its product is equal to the product of first and last term.
$ = {x^2} + 2x + 5x + 10 $
Make the pair of first two terms and the last two terms.
$ = \underline {{x^2} + 2x} + \underline {5x + 10} $
Take common factors from the paired terms in the above expression.
$ = x(x + 2) + 5(x + 2) $
Take common factor common from the above expression –
$ = (x + 2)(x + 5) $
Here we will get two zeroes –
$ x + 2 = 0 $
Make the required term “x” the subject and move constantly on the opposite side. When you move any term from one side to another then the sign of the term changes. Positive term becomes negative and vice-versa.
$ \Rightarrow x = - 2 $ ….. (A)
Now, $ x + 5 = 0 $
Make “x” the subject –
$ \Rightarrow x = - 5 $ ….. (B)
Now, the sum of zeroes of the given quadratic polynomials are given by $ = - 2 - 5 $
Simplify combining the two zeros
Sum of zeroes $ = \dfrac{{ - 7}}{1} = \dfrac{{ - x\,{\text{coefficient }}}}{{{x^2}coefficient}} $
Now, the product of roots can be expressed as $ = ( - 2)( - 5) $
Simplify the expression finding the product of two negative terms which gives positive terms.
Product of roots $ = \dfrac{{10}}{1} = \dfrac{{cons\tan t}}{{{x^2}coefficient}} $
This is the required solution.
Note: Remember that the product of two negative terms gives us the positive term. And when we add or subtract between two terms with different signs do subtraction and apply the sign of the bigger number to the resultant value. Coefficient is the term multiplied with the variable, term is numeric with plus or minus sign.
Complete step-by-step answer:
Take the given expression: $ {x^2} + 7x + 10 $
Split the middle term in such a way that its product is equal to the product of first and last term.
$ = {x^2} + 2x + 5x + 10 $
Make the pair of first two terms and the last two terms.
$ = \underline {{x^2} + 2x} + \underline {5x + 10} $
Take common factors from the paired terms in the above expression.
$ = x(x + 2) + 5(x + 2) $
Take common factor common from the above expression –
$ = (x + 2)(x + 5) $
Here we will get two zeroes –
$ x + 2 = 0 $
Make the required term “x” the subject and move constantly on the opposite side. When you move any term from one side to another then the sign of the term changes. Positive term becomes negative and vice-versa.
$ \Rightarrow x = - 2 $ ….. (A)
Now, $ x + 5 = 0 $
Make “x” the subject –
$ \Rightarrow x = - 5 $ ….. (B)
Now, the sum of zeroes of the given quadratic polynomials are given by $ = - 2 - 5 $
Simplify combining the two zeros
Sum of zeroes $ = \dfrac{{ - 7}}{1} = \dfrac{{ - x\,{\text{coefficient }}}}{{{x^2}coefficient}} $
Now, the product of roots can be expressed as $ = ( - 2)( - 5) $
Simplify the expression finding the product of two negative terms which gives positive terms.
Product of roots $ = \dfrac{{10}}{1} = \dfrac{{cons\tan t}}{{{x^2}coefficient}} $
This is the required solution.
Note: Remember that the product of two negative terms gives us the positive term. And when we add or subtract between two terms with different signs do subtraction and apply the sign of the bigger number to the resultant value. Coefficient is the term multiplied with the variable, term is numeric with plus or minus sign.
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