For the pair of numbers (-1/5),(-1/2) My teacher wants us to find the quadratic equation with integral coefficients that has these numbers as solutions. I am not sure if I have the right equation for this. Could someone let me know if I am right or wrong and then show me how to gettthe right answer if I am wrong. Thanks.
x^2-7/10x+10
x = -1/5 , x = -1/2.
x + 1/5 = 0,
x + 1/2 = 0,
(x + 1/5) (x + 1/2) = 0
Multiply the 2 binomials:
x^2 + x/2 + x/5 + 1/10 = 0
Multiply both sides by the common
denominator:
10x^2 + 5x + 2x + 1 = 0
Combine like-terms:
10x^2 + 7x + 1 = 0
Final Eq: y = 10x^2 + 7x + 1
To determine if you have the correct quadratic equation with integral coefficients, we will check if the given numbers (-1/5) and (-1/2) are indeed solutions to the equation. To do this, we substitute each value into the equation and check if the equation holds true.
Let's test the first solution, (-1/5):
Substituting x = -1/5 into the equation, we get:
((-1/5)^2) - (7/10)(-1/5) + 10 = 1/25 + 7/50 + 10 = 2/50 + 7/50 + 10 = 9/50 + 10
This is not equal to zero, which means (-1/5) is not a solution of the equation.
Now, let's test the second solution, (-1/2):
Substituting x = -1/2 into the equation, we get:
((-1/2)^2) - (7/10)(-1/2) + 10 = 1/4 + 7/20 + 10 = 5/20 + 7/20 + 10 = 12/20 + 10 = 30/20 + 10 = 3/2 + 10 = 13/2
This is not equal to zero, which means (-1/2) is also not a solution of the equation.
Since both given numbers are not solutions to the equation x^2 - (7/10)x + 10, the equation you provided is incorrect.
To find the correct quadratic equation with integral coefficients using the given solutions, we will apply Vieta's formulas. Vieta's formulas relate the coefficients of a quadratic equation to the roots, or solutions, of the equation.
Let's use the given solutions (-1/5) and (-1/2) to form the quadratic equation:
x = (-1/5) or x = (-1/2)
To get rid of the denominators, multiply both sides of each equation by a common multiple of the denominators. In this case, the least common multiple (LCM) of 5 and 2 is 10:
10x = -2 or 10x = -5
Now we need to eliminate the decimal coefficients to obtain an equation with integral coefficients. Multiply both sides of each equation by 10:
10x^2 = -20 or 10x^2 = -50
The resulting quadratic equations that have (-1/5) and (-1/2) as solutions are:
10x^2 + 20 = 0 or 10x^2 + 50 = 0
After simplifying, the equation with integral coefficients for the given solutions (-1/5) and (-1/2) is:
10x^2 + 20 = 0 or 10x^2 + 50 = 0