Question

2.

In this question you must show all steps of your working.
Solutions relying on calculator technology are not acceptable.
#
Figure 1
Figure 1 shows a sketch of the curve with equation y = 2r² + 3x and the straight line with
1
equation y=x+.
2x + 3
The line meets the curve at the points P and Q, as shown in Figure 1.
(a) Using algebra, find the coordinates of P and the coordinates of Q.
(b) Hence write down the range of values of x for which 2r² + 3.x >
1|2
+3
(5)
(2)
blank

Answer 1

`\sf a) \quad P=(-2, 2) \: \textsf{ and }\: \sf Q= \left((3)/(4), (27)/(8)\right)`

`\textsf{b)} \quad x\leq -2 \:\:\: \textsf {or }\:\:x\geq (3)/(4)`

Explanation:

Part (a)

Given equations:

`\begin{cases}y=(1)/(2)x+3\\\\y=2x^2+3x\end{cases}`

Points P and Q are the points of intersection of the given equations.

Therefore, to find the points of intersection, equate the equations and solve for x:

`\implies 2x^2+3x=(1)/(2)x+3`

`\implies 2x^2+(5)/(2)x-3=0`

`\implies 2\left(2x^2+(5)/(2)x-3\right)=2(0)`

`\implies 4x^2+5x-6=0`

`\implies 4x^2+8x-3x-6=0`

`\implies 4x(x+2)-3(x+2)=0`

`\implies (4x-3)(x+2)=0`

Apply the zero-product property:

`\implies 4x-3=0 \implies x=(3)/(4)`

`\implies x+2=0 \implies x=-2`

Substitute the found values of x into one of the equations to find the y-coordinates of points P and Q:

`x_P=-2 \implies y_P=(1)/(2)(-2)+3=2`

`x_Q=(3)/(4) \implies y_Q=(1)/(2)\left((3)/(4)\right)+3=(27)/(8)`

Therefore, the coordinates of P and Q are:

`\sf P=(-2, 2) \: \textsf{ and }\: \sf Q= \left((3)/(4), (27)/(8)\right)`

Part (b)

Given inequality:

`2x^2+3x \geq (1)/(2)x+3`

The range of values that satisfies the given inequality is the range of x-values where the curve is equal to or higher than the line.

The curve is equal to or higher than the line for x-values equal to or less than point P and x-values equal to or more than point Q:

`x\leq -2 \:\:\: \textsf {or }\:\:x\geq (3)/(4)`