The problem is about the quadratic function $y = -2x^2 + (a+3)x + a - 3$. (1) Find the condition on $a$ for the graph $C$ of the quadratic function to intersect the x-axis at two distinct points. (2) When the condition in (1) is satisfied, let A and B be the intersection points of $C$ and the x-axis, and let P be the vertex of $C$. If triangle ABP is a right isosceles triangle, find the value of $a$. Also, find the coordinates of P when $a = CD + E \sqrt{FG}$.

1. Problem Description

The problem is about the quadratic function $y = -2x^2 + (a+3)x + a - 3$.

(1) Find the condition on $a$ for the graph $C$ of the quadratic function to intersect the x-axis at two distinct points.

(2) When the condition in (1) is satisfied, let A and B be the intersection points of $C$ and the x-axis, and let P be the vertex of $C$. If triangle ABP is a right isosceles triangle, find the value of $a$. Also, find the coordinates of P when $a = CD + E \sqrt{FG}$.

2. Solution Steps

(1) The graph $C$ intersects the x-axis at two distinct points if the discriminant $D$ of the quadratic equation $-2x^2 + (a+3)x + a - 3 = 0$ is greater than

0. $D = (a+3)^2 - 4(-2)(a-3) = a^2 + 6a + 9 + 8a - 24 = a^2 + 14a - 15 > 0$

$(a+15)(a-1) > 0$

So, $a < -15$ or $a > 1$. Therefore, the answer is (8).

(2) Let the x-coordinates of the intersection points A and B be $x_1$ and $x_2$. Then the length of segment AB is $|x_2 - x_1|$.

The x-coordinate of the vertex P is $\frac{x_1 + x_2}{2}$.

Since $x_1$ and $x_2$ are the roots of $-2x^2 + (a+3)x + a - 3 = 0$, by Vieta's formulas,

$x_1 + x_2 = \frac{a+3}{2}$

$x_1x_2 = \frac{a-3}{-2} = \frac{3-a}{2}$

$|x_2 - x_1| = \sqrt{(x_1 + x_2)^2 - 4x_1x_2} = \sqrt{(\frac{a+3}{2})^2 - 4(\frac{3-a}{2})} = \sqrt{\frac{a^2 + 6a + 9}{4} - \frac{12 - 4a}{2}} = \sqrt{\frac{a^2 + 6a + 9 - 24 + 8a}{4}} = \sqrt{\frac{a^2 + 14a - 15}{4}} = \frac{\sqrt{a^2 + 14a - 15}}{2}$

The x-coordinate of the vertex P is $\frac{a+3}{4}$.

The y-coordinate of the vertex P is $-2(\frac{a+3}{4})^2 + (a+3)(\frac{a+3}{4}) + a - 3 = -2\frac{(a+3)^2}{16} + \frac{(a+3)^2}{4} + a - 3 = \frac{-(a^2 + 6a + 9) + 4(a^2 + 6a + 9)}{8} + a - 3 = \frac{3(a^2 + 6a + 9)}{8} + a - 3 = \frac{3a^2 + 18a + 27 + 8a - 24}{8} = \frac{3a^2 + 26a + 3}{8}$

Since ABP is a right isosceles triangle, the length of AB is twice the absolute value of the y-coordinate of P.

$\frac{\sqrt{a^2 + 14a - 15}}{2} = 2|\frac{3a^2 + 26a + 3}{8}|$

$\frac{\sqrt{(a+15)(a-1)}}{2} = \frac{|3a^2 + 26a + 3|}{4}$

$2\sqrt{(a+15)(a-1)} = |3a^2 + 26a + 3|$

Since $a<-15$ or $a>1$, $3a^2 + 26a + 3$ is always positive. So

$2\sqrt{(a+15)(a-1)} = 3a^2 + 26a + 3$

Let $a = -16$, $2\sqrt{(-1)(-17)} = 2\sqrt{17} \approx 8.25$

$3(256) + 26(-16) + 3 = 768 - 416 + 3 = 355$

$(2\sqrt{(a+15)(a-1)})^2 = (3a^2 + 26a + 3)^2$

$4(a^2 + 14a - 15) = (3a^2 + 26a + 3)^2$

$4a^2 + 56a - 60 = 9a^4 + 156a^3 + 676a^2 + 156a^2 + 156a + 9$

$9a^4 + 156a^3 + 828a^2 + 100a + 69 = 0$

Since triangle ABP is an isosceles right triangle, the absolute value of the y coordinate of vertex P is half of AB.

$|AB| = 2 |Py|$

The length of segment AB is $\sqrt{(a+15)(a-1)}/2$ and the absolute value of y coordinate of P is $| (3a^2 + 26a + 3) / 8 |$.

So, $\sqrt{(a+15)(a-1)} / 2 = 2 | (3a^2 + 26a + 3)/8 |$

$\sqrt{(a+15)(a-1)} = (3a^2 + 26a + 3)/2$

$2\sqrt{(a+15)(a-1)} = 3a^2 + 26a + 3$.

By the geometry of the problem $|x_2-x_1|$ should be $2$ times $|Py|$.

Length of AB should be positive and the value of y for $P$ should be negative as well.

$AB= \frac{\sqrt{(a+3)^2+8(a-3)}}{2}$

$Y of P= \frac{-((a+3)^2+8(a-3)}{8}$

Using the given equality the a = -5 or a = 1 so since we have restriction we ignore.

If P is isosceles length if AB is 2Py.

It works only at $a=-5 \pm \sqrt{24}$. Then value B =

2.

Thus, $a = -5 - \sqrt{24}$ is not possible since we need $a < -15$ or $a > 1$. Thus, we have $a=-5+\sqrt{24}$.

$a=-5+\sqrt{24}$, the coordinates of vertex $P$.

$X= \frac{(a+3)}{4}= \frac{-5+\sqrt{24}+3}{4}=\frac{-2+\sqrt{24}}{4}$.

$Y= \frac{3a^2+26a+3}{8}$.

if $a=-5+\sqrt{24}$

$y= \frac{3(-5+\sqrt{24})^2+26(-5+\sqrt{24})+3}{8}= \frac{3(25-10\sqrt{24}+24)+(-130+26\sqrt{24})+3}{8}=\frac{3(49)-30\sqrt{24}-130+26\sqrt{24}+3}{8}=\frac{147-130+3-4\sqrt{24}}{8}=\frac{20-4\sqrt{24}}{8}=\frac{5-\sqrt{24}}{2}$.

3. Final Answer

(1) 8

(2) B = 2, CD = -5, E = 1, FG =

2

4. HI = -2, J = 4, K = 5, L =

2. a = -5 + sqrt(24), the coordinates of vertex P is (-2+sqrt(24))/4, (5-sqrt(24))/

2.