When solving geometry problems with constraints, carefully analyze integer solutions and relationships between coordinates
The intercepts of the line are (20,0) and (0,15). Compute the points inside the triangle satisfying b = ka. The total is: 31points.
The correct answer is 31
\((1,1)(1,2)−(1,14)⇒14 pts.\)
If \(x=2\), \(y=\frac{27}{2} = 13.5\)
\((2,2)(2,4)…(2,12)⇒6 pts.\)
If \(x=3\), \(y=\frac{51}{4} = 12.75\)
\((3,3)(3,6)−(3,12)⇒4 pts.\)
If \(x=4\), \(y=12\)
\((4,4)(4,8)⇒2 pts.\)
If \(x=5\), \(y=\frac{45}{4} = 11.25\)
\((5,5),(5,10)⇒2 pts.\)
If \(x=6\), \(y=\frac{21}{2} = 10.5\)
\((6,6)⇒1pt\)
If \(x=7\), \(y=\frac{39}{4} = 9.75\)
(7,7)⇒1pt
If \(x=8\), \(y=9\)
\((8,8)⇒1pt\)
If \(x=9\) \(y=\frac{33}{4}= 8.25\)\(\Rightarrow\) no pt.
Total = 31 pts.
If the roots of the quadratic equation \( ax^2 + bx + c = 0 \) are real and equal, then:
A polynomial that has two roots or is of degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b, and c are the real numbers.
Consider the following equation ax²+bx+c=0, where a≠0 and a, b, and c are real coefficients.
The solution of a quadratic equation can be found using the formula, x=((-b±√(b²-4ac))/2a)
Read More: Nature of Roots of Quadratic Equation