Solve by factoring
x^2+4x+5=0
need help
It has no real roots. Should the +5 be -5?
(x-1)(x+5)
for x^2+4x-5
I wrote is down wrong. Here is the correct problem.
Solving by factoring
X^2+4x+3=0
To solve this quadratic equation by factoring, you need to find two numbers that satisfy two conditions:
1. The product of the two numbers should be equal to the constant term (in this case, 5).
2. The sum of the two numbers should be equal to the coefficient of the linear term (in this case, 4).
Let's go through the steps to solve the equation:
1. Write down the equation: x^2 + 4x + 5 = 0
2. Look for two numbers that satisfy the conditions mentioned above. In this case, you need to find two numbers whose product is 5 and whose sum is 4.
The possible pairs of factors for 5 are (1, 5) and (-1, -5).
3. Rewrite the middle term (4x) using the factors you found in step 2. Replace the 4x term using the factors that have a sum of 4x. Since negative 1 + negative 5 equals negative 6, that gives us:
x^2 - x - 5x + 5 = 0
4. Group the terms and factor by grouping. Group the first two terms and the last two terms separately and factor out the common factors.
(x^2 - x) + (-5x + 5) = 0
Factor out x from the first group and -5 from the second group:
x(x - 1) - 5(x - 1) = 0
5. Notice that you now have a common binomial factor, (x - 1), in both terms.
Combine the two terms with the common factor:
(x - 1)(x - 5) = 0
6. Now, set each factor equal to zero and solve for x separately.
x - 1 = 0 --> x = 1
x - 5 = 0 --> x = 5
7. The solutions to the equation x^2 + 4x + 5 = 0 are x = 1 and x = 5.
By factoring the quadratic equation, we found that x can take on two values, 1 and 5, which satisfy the equation.