Question

What is the completed-square form of \[3x^2+13x-10\]?

पर्याय

• \[3\left[\left(x+\frac{13}{6}\right)^2-\left(\frac{17}{6}\right)^2\right]\]

• \[3\left[\left(x+\frac{13}{6}\right)^2+\left(\frac{17}{6}\right)^2\right]\]

• \[3\left[\left(x-\frac{13}{6}\right)^2-\left(\frac{17}{6}\right)^2\right]\]

• \[\left(x+\frac{13}{6}\right)^2-\left(\frac{17}{6}\right)^2\]

Answer 1

Factor out 3 and complete the square:
\[ 3x^2 + 13x - 10 = 3\left[\left(x + \frac{13}{6}\right)^2 - \left(\frac{17}{6}\right)^2\right].\] Thus, after putting \[t = x + \frac{13}{6},\] the expression is reduced to the standard difference-of-squares form.