Asked by Mildred Oshio
Find the quadratic equation whose roots are given -2/3 and -1/4 with shown working below
Answers
Answered by
Bot
The quadratic equation is given by ax^2 + bx + c = 0, where a, b and c are constants.
We can use the formula for finding the quadratic equation given the roots:
a = 1, b = (root1 + root2), c = root1 * root2
Therefore, the quadratic equation is:
x^2 + ( (-2/3) + (-1/4) )x + (-2/3) * (-1/4) = 0
x^2 - 5/12x - 1/12 = 0
We can use the formula for finding the quadratic equation given the roots:
a = 1, b = (root1 + root2), c = root1 * root2
Therefore, the quadratic equation is:
x^2 + ( (-2/3) + (-1/4) )x + (-2/3) * (-1/4) = 0
x^2 - 5/12x - 1/12 = 0
Answered by
Mildred Oshio
The quadratic equation is given by ax^2 + bx + c = 0, where a, b and c are constants.
We can use the formula for finding the quadratic equation given the roots:
a = 1, b = (root1 + root2), c = root1 * root2
Therefore, the quadratic equation is:
x^2 + ( (-2/3) + (-1/4) )x + (-2/3) * (-1/4) = 0
x^2 - 5/12x - 1/12 = 0
We can use the formula for finding the quadratic equation given the roots:
a = 1, b = (root1 + root2), c = root1 * root2
Therefore, the quadratic equation is:
x^2 + ( (-2/3) + (-1/4) )x + (-2/3) * (-1/4) = 0
x^2 - 5/12x - 1/12 = 0
Answered by
oobleck
AAAaannndd the bot gets it wrong yet again!
TWICE!
(3x+2)(4x+1) = 0
TWICE!
(3x+2)(4x+1) = 0
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