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kaliprasad
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form a quadratic equation $ax^2 + bx + c = 0$ such that a,b,c are in AP and it has got integer roots
[sp]In fact, assuming that $a$, $b$ and $c$ are integers there is only one other solution apart from $x^2 + 8x + 15 = 0$ or multiples of it.Bacterius said:[sp]Set $a = 1$ for convenience. Then the quadratic has integer roots if and only if $b^2 - 4c$ is a perfect square. Furthermore we need $a$, $b$ and $c$ to be in arithmetic progression, so we must have $b = 1 + k$ and $c = 1 + 2k$ for some nonzero integer $k$. Thus we need $(1 + k)^2 - 4(1 + 2k) = k^2 - 6k - 3$ to be a perfect square. The first one that works is $k = 7$, since $7^2 - 6 \times 7 - 3 = 4 = 2^2$. Therefore a solution is:[/sp]
$$x^2 + 8x + 15 = 0$$
Which has integer roots $-5$ and $-3$, and $1$, $8$, $15$ are in arithmetic progression.
TODO: find a systematic way to choose $k$, and prove/disprove there are infinitely many solutions for any $a \ne 0$.
A quadratic equation is a mathematical equation in the form of ax^2 + bx + c = 0, where a, b, and c are constants and x is the variable. It is also known as a second degree polynomial equation.
There are several methods for solving a quadratic equation, including factoring, using the quadratic formula, and completing the square. The appropriate method to use depends on the specific equation and its coefficients.
The discriminant of a quadratic equation is the part of the equation under the square root sign in the quadratic formula, b^2 - 4ac. It helps determine the nature of the solutions of the equation. If the discriminant is positive, there are two real solutions. If it is zero, there is one real solution. If it is negative, there are no real solutions.
Yes, a quadratic equation can have imaginary solutions if the discriminant is negative. In this case, the solutions will be in the form of complex numbers, with a real and imaginary part.
Quadratic equations can be used to model various real-life situations, such as the path of a projectile, the shape of a parabolic antenna, or the profit and loss in business. They can also be used to solve optimization problems, such as finding the maximum or minimum value of a function.