Concept explainers
(a)
To find:the roots of the equation using middle term factorization.
(a)
Answer to Problem 123E
The solution of the given equation
Explanation of Solution
Given:
Concept used:
The First it can solve the given equation through middle term factorization by taking common factor and if it doesn’t get factorize then use discriminant method to solve the given quadratic equation where
Calculation:
First by solving the equation through middle term factorization
If it doesn’t get factorized then it can be solved from discriminant method to solve the given quadratic equation.
Where the equation is defined by
Where s =sum of the two roots and p= product of two roots
Let the two roots be
Sum of the roots =
Here in given solution
Here need to find 2 number whose sum is
Which is possible by
So, two numbers are
Since,
So,
Hence, the solution of the given equation
(b)
To find:the roots of the equation using middle term factorization or by quadratic formula.
(b)
Answer to Problem 123E
The solution of
Explanation of Solution
Given:
Concept used:
The First it can solve the given equation through middle term factorization by taking common factor and if it doesn’t get factorize then use discriminant method to solve the given quadratic equation where
Calculation:
Here the equation is defined by
Where s =sum of the two roots and p= product of two roots
Let the two roots be
Sum of the roots =
Here in given solution
This is not possible since sum of the root and product of the root is not equal so it cannot be solved from middle term factorization method.
It can be solved from discriminant method:
Here
Hence the solution of
(c)
To prove:that the quadratic equation has two roots which when multiply give product of the root C and which when sum give sum of the root B.
(c)
Answer to Problem 123E
It is true that two roots which when multiply give product of the root C and which when sum give sum of the root B.
Explanation of Solution
Given:
Concept used:
Using discriminant method to solve the given quadratic equation
where
Calculation:
Chapter 1 Solutions
Precalculus: Mathematics for Calculus - 6th Edition
- Calculus: Early TranscendentalsCalculusISBN:9781285741550Author:James StewartPublisher:Cengage LearningThomas' Calculus (14th Edition)CalculusISBN:9780134438986Author:Joel R. Hass, Christopher E. Heil, Maurice D. WeirPublisher:PEARSONCalculus: Early Transcendentals (3rd Edition)CalculusISBN:9780134763644Author:William L. Briggs, Lyle Cochran, Bernard Gillett, Eric SchulzPublisher:PEARSON
- Calculus: Early TranscendentalsCalculusISBN:9781319050740Author:Jon Rogawski, Colin Adams, Robert FranzosaPublisher:W. H. FreemanCalculus: Early Transcendental FunctionsCalculusISBN:9781337552516Author:Ron Larson, Bruce H. EdwardsPublisher:Cengage Learning