Answer:
To write a quadratic equation in standard form with the given roots of \( \frac{3}{4} \) and -5, you can follow these steps:
1. Start with the factored form of a quadratic equation, which is \(a(x - r_1)(x - r_2) = 0\), where \(a\) is the leading coefficient and \(r_1\) and \(r_2\) are the roots.
2. Plug in the given roots:
\(a\left(x - \frac{3}{4}\right)(x + 5) = 0\)
3. To find the value of \(a\), you can use any point on the quadratic curve. You have two roots, so you can choose either one. Let's use the root \(x = \frac{3}{4}\). Plug in this value and solve for \(a\):
\(a\left(\frac{3}{4} - \frac{3}{4}\right)\left(\frac{3}{4} + 5\right) = 0\)
\(a(0)\left(\frac{27}{4}\right) = 0\)
Since the left side is equal to 0, we can conclude that \(a\) can be any real number.
4. Now, you have your quadratic equation in factored form:
\(a\left(x - \frac{3}{4}\right)(x + 5) = 0\)
5. If you want the quadratic equation in standard form (ax^2 + bx + c = 0), you can expand the equation:
\(a(x - \frac{3}{4})(x + 5) = 0\)
\(a(x^2 + 5x - \frac{3}{4}x - \frac{15}{4}) = 0\)
\(a(x^2 + \frac{17}{4}x - \frac{15}{4}) = 0\)
Now, your quadratic equation in standard form with the given roots \( \frac{3}{4} \) and -5 is:
\(a(x^2 + \frac{17}{4}x - \frac{15}{4}) = 0\)
You can multiply all terms by a constant if you want a specific value for \(a\), but in general, this is the standard form of the quadratic equation with the given roots.