Question

What value of \[b\] would make \[{y^2} + by - 24\] factorable?

Answer 1

Hint: We have to solve the quadratic equation by finding out such a value that the given equation becomes factorable. We can use the prime factorization method to split up the equation and find out the answer to the coefficient of \[y\].

Complete step by step solution:
Quadratic equations are the polynomial equations of degree \[2\] in one variable of type \[f(x) = a{x^2} + bx + c\] where \[a,b,c \in R\] and \[a \ne 0\]. It is the general form of a quadratic equation where ‘\[a\]’ is called the leading coefficient and ‘\[c\]’ is called the absolute term of \[f(x)\].
The quadratic equation will always have two roots.
Factors are the numbers which divide the given number exactly, whereas the multiples are the numbers which are multiplied by the other number to get specific numbers. For example- factors of \[10\] are \[1,2,5,10\] and their additive inverses i.e. \[ - 1, - 2, - 5, - 10\].
We can solve the given sum as follows:
We know that in \[a{x^2} + bx + c\], as per factorization method, the second term \[bx\] is the sum of the factors where the last term \[c\] is the multiplication of factors.
Now we can write \[24\] as multiplication of factors in \[{y^2} + by - 24\] as follows:
\[24 = 8 \times 3\]
\[24 = 12 \times 2\]
\[24 = 4 \times 6\]
\[24 = 24 \times 1\]
Since we have \[ - 24\], either of the terms of multiplication can be negative.
Hence the possible values of \[b\] can be as follows after subtracting the factors found above:
\[ b = - (8 - 3) = - 5\]
\[b = - (12 - 2) = - 10\]
\[b = - (6 - 4) = - 2\]
\[b = - (24 - 1) = - 23\]
Therefore, the value of \[b\] can be \[ - 2, - 5, - 10, - 23\] or \[2,5,10,23\] since additive inverses are also included. This will make the equation \[{y^2} + by - 24\] factorable.

Note:
We can use the discriminant of the quadratic equation \[{b^2} - 4ac\] to find out whether the equation is factorable or not. The equation will only be factorable if discriminant is a perfect square. In the given case,
\[{b^2} - 4(1)( - 24)\]
\[{b^2} + 96 > 0\] (always positive since square of any number cannot be negative)
Hence, rational and distinct roots exist for \[{y^2} + by - 24\].