x2+bx+75
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Solve B
I assume that you want to solve for the value of b in the quadratic equation x^2 + bx + 75 = 0.
Using the quadratic formula,
x = [-b ± sqrt(b^2 - 4ac)]/2a
where a = 1, b = b, and c = 75.
For the equation to have real solutions, the discriminant (b^2 - 4ac) must be greater than or equal to zero.
So, we have:
b^2 - 4ac >= 0
b^2 - 4(1)(75) >= 0
b^2 - 300 >= 0
b^2 >= 300
b >= sqrt(300) or b <= -sqrt(300)
Therefore, the solution for b is:
b >= sqrt(300) or b <= -sqrt(300)
Approximating the value of sqrt(300) to the nearest tenth gives:
b >= 17.3 or b <= -17.3
So the value of b must be greater than or equal to 17.3 or less than or equal to -17.3 for the quadratic equation x^2 + bx + 75 = 0 to have real solutions.
To factor the quadratic expression \(x^2+bx+75\), we are looking for two binomials in the form \( (x+m)(x+n) \), where \(m\) and \(n\) are constants.
The constant terms of the binomials must multiply to give \(75\). So, we need to find two numbers that multiply to 75. The possible pairs of factors of 75 are:
1, 75
3, 25
5, 15
Next, we look for the pair of factors that, when added together, give the coefficient of the \(x\) term (\(b\)).
For example, if \(b = 7\), the pair of factors 3 and 25 would be appropriate because \(3 + 25 = 28\), which matches the coefficient \(b\).
In this case, we don't have a specific value for \(b\), so we need to keep the factors in the pairs as variables. Let's use \(m\) and \(n\) as the variables.
The factorization could then be written as:
\(x^2+bx+75 = (x+m)(x+n)\)
Where \(m\) and \(n\) are two numbers that multiply to give 75 and add up to \(b\).