sum and product It is a method used to find the solutions of a equation. We use the sum and product as a method to calculate the roots of a 2nd degree equation, of the type ax² + bx + c = 0.
This is an interesting method when the solutions of the equation are whole numbers. In cases where the solutions are not integers, it can be quite complicated to use the sum and product, with other easier methods to find the solutions of the equation.
Read too: Bhaskara — the best-known formula for solving quadratic equations
Topics of this article
- 1 - Summary about sum and product
- 2 - What is the sum and product?
- 3 - Sum and product formula
- 4 - How to calculate the roots using sum and product?
- 5 - Solved exercises on sum and product
Summary about sum and product
- The sum and product is one of the methods used to find the solutions of a complete quadratic equation.
- By the sum and product, given the equation of the 2nd degree ax² + bx + c = 0, we have:
\(x_1+x_2=-\frac{b}{a}\)
\(x_1\cdot x_2=\frac{c}{a}\)
- x1 It is x2 are the solutions of the quadratic equation.
- a, b and c are the coefficients of the 2nd degree equation.
What is sum and product?
The sum and product is one of the methods we can use to find the solutions of an equation. Used in 2nd degree equations, the sum and product can be a more practical method to find the solutions of the equation, because it consists of looking for the numbers that satisfy the sum and product formula for a given equation.
Sum and product formula
In a quadratic equation, of the type ax² + bx + c = 0, with solutions equal to x1 and x2, by sum and product, we have:
\(x_1+x_2=-\frac{b}{a}\)
\(x_1\cdot x_2=\frac{c}{a}\)
Do not stop now... There's more after the publicity ;)
How to calculate roots using sum and product?
To find the solutions, we first look for the integers whose product is equal to \(\frac{c}{a}\).
We know that the solutions of the equation can be positive or negative:
- Positive product and positive sum: both roots are positive.
- Positive product and negative sum: both roots are negative.
- Negative product and positive sum: one root is positive and the other is negative, and the one with the largest module is positive.
- Negative product and negative sum: one root is positive and the other is negative, and the one with the largest module is negative.
Later, after listing all the products that satisfy the equation, we analyze which one satisfies the equation. equation of the sum, that is, what are the two numbers that satisfy the equation of the product and the sum simultaneously.
Example 1:
Find the solutions of the equation:
\(x²-5x+6=0\)
At first, we'll substitute into the sum and product formula. We have that a = 1, b = -5 and c = 6:
\(x_1+x_2=5\)
\(x_1\cdot x_2=6\)
Since the sum and product are positive, the roots are positive. Analyzing the product, we know that:
\(1\ \cdot6\ =\ 6\ \)
\(2\cdot3\ =\ 6\)
Now, we will check which of these results has a sum equal to 5, which in this case is:
\(2+3=5\)
So, the solutions of this equation are \(x_1=2\ and\ x_2=3\).
Example 2:
Find the solutions of the equation:
\(x^2+2x-24=0\ \)
First, we'll substitute into the sum and product formula. We have a = 1, b = 2 and c = -24.
\(x_1+x_2=-\ 2\)
\(x_1\cdot x_2=-\ 24\)
Since the sum and product are negative, the roots are of opposite signs, and the one with the greatest modulus is negative. Analyzing the product, we know that:
\(1\cdot(-24)=-24\)
\(2\cdot\left(-12\right)=-24\)
\(3\cdot\left(-8\right)=-24\)
\(4\cdot\left(-6\right)=-24\)
Now, let's check which of these results has a sum equal to -2, which in this case is:
\(4+\left(-6\right)=-2\)
So, the solutions of this equation are \(x_1=4\ and\ x_2=-6\) .
Read too: How to solve an incomplete quadratic equation
Solved exercises on sum and product
question 1
be y It is z the roots of equation 4x2-3x-1=0, the value of 4(y+4)(z+4) é:
A) 75
B) 64
C) 32
D) 18
E) 16
Resolution:
Alternative A
Calculating by sum and product:
\(y+z=\frac{3}{4}\)
\(y\cdot z=-\frac{1}{4}\)
So, we have to:
\(4\left (y+4\right)\left (z+4\right)=4(yz+4y+4z+16)\)
\(4\left (y+4\right)\left (z+4\right)=4\left(-\frac{1}{4}+4\left (y+z\right)+16\right )\)
\(4\left (y+4\right)\left (z+4\right)=4\left(-\frac{1}{4}+4\cdot\frac{3}{4}+16\ right)\)
\(4\left (y+4\right)\left (z+4\right)=4\left(-\frac{1}{4}+3+16\right)\)
\(4\left (y+4\right)\left (z+4\right)=4\left(-\frac{1}{4}+19\right)\)
\(4\left (y+4\right)\left (z+4\right)=4\left(\frac{76-1}{4}\right)\)
\(4\left (y+4\right)\left (z+4\right)=4\cdot\frac{75}{4}\)
\(4\left (y+4\right)\left (z+4\right)=75\)
question 2
Considering the equation 2x2 + 8x + 6 = 0, let S be the sum of the roots of this equation and P be the product of the roots of the equation, then the value of the operation (S - P)2 é:
A) 36
B) 49
C) 64
D) 81
E) 100
Resolution:
Alternative B
Calculating by sum and product:
\(S=x_1+x_2=-4\)
\(P\ =\ x_1\cdot x_2=3\)
So, we have to:
\(\left(-4-3\right)^2=\left(-7\right)^2=49\)
By Raul Rodrigues de Oliveira
Maths teacher
Would you like to reference this text in a school or academic work? Look:
OLIVEIRA, Raul Rodrigues de. "Sum and Product"; Brazil School. Available in: https://brasilescola.uol.com.br/matematica/soma-e-produto.htm. Accessed on July 22, 2023.
Click to see the demonstration of Bhaskara's formula, which is based on the completing square method.
Understand what a 2nd degree equation is. Learn how to calculate your roots and Bhaskara's formula. Also learn how to solve a system of 2nd degree equations.
Learn what it is and how to use Bhaskara's formula to solve quadratic equations!
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