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Quadratic Equations
Posted by Shriya on June 24, 2023 at 8:13 pmIf the quadratic equation x<sup>2</sup> + bx + 72 = 0 has 2 distinct
integer roots then the number of possible values for b isKomal Bisht replied 1 year, 2 months ago 3 Members · 2 Replies -
2 Replies
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Given quadratic: x^2 + bx + 72 = 0 has 2 distinct integer roots.
Then, the determinant b^2 – 4ac >0 for standard equation:- ax^2 + bx+ c
Here a = 1, b = b, c = 72
So, b^2 – 4.1.72 >0
= b^2 > 288
Now the least number greater than 288 which is a perfect square is 289. So b >= 17
This ensures the root is real and distinct.
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To find the number of possible values for b in the quadratic equation x^2 + bx + 72 = 0, which has 2 distinct integer roots, we can use the fact that the product of the roots of a quadratic equation is equal to the constant term divided by the coefficient of the squared term. Also the sum of roots is equal to the coefficient of x which is -b/x^2.
In this case, the constant term is 72 and the coefficient of the squared term is 1. Therefore, the product of the roots is 72.
To have 2 distinct integer roots, the product of the roots must be factorable into two different pairs of integers whose difference is not zero.
The factors of 72 are: 1, 72 2, 36 3, 24 4, 18 6, 12 8, 9.
Therefore, there are 6 possible values for b in the given quadratic equation.
Pair (1, 72): For this pair, the sum of the roots is 1 + 72 = 73. Since the sum of the roots of a quadratic equation is equal to the negation of the coefficient of the linear term (b), we have b = -73.
Pair (2, 36): The sum of the roots is 2 + 36 = 38, so b = -38.
Pair (3, 24): The sum of the roots is 3 + 24 = 27, so b = -27.
Pair (4, 18): The sum of the roots is 4 + 18 = 22, so b = -22.
Pair (6, 12): The sum of the roots is 6 + 12 = 18, so b = -18.
Pair (8, 9): The sum of the roots is 8 + 9 = 17, so b = -17.
Therefore, the possible values for b in the given quadratic equation x^2 + bx + 72 = 0, with 2 distinct integer roots, are: b = -73, -38, -27, -22, -18, -17.
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