Determine the zero(s) of the quadratic function y = 6x^ + 13x + 5. Show all steps.
The quadratic function is y = 6x^2 + 13x + 5.
To find the zero(s) of the quadratic function, we set y equal to zero and solve for x:
0 = 6x^2 + 13x + 5
To factor this quadratic equation, we need to find two numbers whose sum is 13 and whose product is (6)(5) = 30. The numbers are 10 and 3, since 10 + 3 = 13 and 10 * 3 = 30.
0 = 6x^2 + 10x + 3x + 5 (applying the numbers 10 and 3 as the coefficients of x)
0 = (6x^2 + 10x) + (3x + 5)
Now, we factor the equation by grouping:
0 = 2x(3x + 5) + 1(3x + 5)
0 = (2x + 1)(3x + 5)
Now, we can set each factor equal to zero:
2x + 1 = 0 or 3x + 5 = 0
Solving the first equation, we get:
2x = -1
x = -1/2
Solving the second equation, we get:
3x = -5
x = -5/3
Therefore, the zeros of the quadratic function y = 6x^2 + 13x + 5 are x = -1/2 and x = -5/3.
To find the zeros of the quadratic function y = 6x^2 + 13x + 5, we need to find the values of x that make y equal to zero.
Step 1: Set y to zero:
0 = 6x^2 + 13x + 5
Step 2: Factor the quadratic equation:
The equation doesn't factor easily, so we can use the quadratic formula to find the zeros.
The quadratic formula is given by:
x = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 6, b = 13, and c = 5.
Step 3: Substitute the values into the quadratic formula:
x = (-13 ± √(13^2 - 4(6)(5))) / 2(6)
Step 4: Simplify the expression inside the square root:
x = (-13 ± √(169 - 120)) / 12
x = (-13 ± √49) / 12
x = (-13 ± 7) / 12
Step 5: Calculate the two possible solutions:
x1 = (-13 + 7) / 12
x1 = -6 / 12
x1 = -0.5
x2 = (-13 - 7) / 12
x2 = -20 / 12
x2 = -1.67 (rounded to two decimal places)
Therefore, the zeros of the quadratic function y = 6x^2 + 13x + 5 are x = -0.5 and x = -1.67.
To determine the zeros of a quadratic function, we need to find the values of x that make the function equal to zero. In other words, we need to solve the equation 6x^2 + 13x + 5 = 0.
To solve this equation, we can use the quadratic formula, which states that for a quadratic equation of the form ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 6, b = 13, and c = 5. Plugging these values into the quadratic formula, we get:
x = (-13 ± √((13^2) - 4(6)(5))) / (2(6))
Let's simplify this expression step by step:
x = (-13 ± √(169 - 120)) / 12
x = (-13 ± √49) / 12
x = (-13 ± 7) / 12
Now we have two possible solutions:
x1 = (-13 + 7) / 12 = -6 / 12 = -1/2
x2 = (-13 - 7) / 12 = -20 / 12 = -5/3
Therefore, the zeros of the given quadratic function y = 6x^2 + 13x + 5 are x = -1/2 and x = -5/3.