Hello,
I am having trouble with Questions 7 and 10! Can someone please help?
Thanks :)
https://www.examinations.ie/schools/Mathematics_Sample_Paper_P1__Higher_Level.pdf

Dazzla16
Hi RyanS! Here are the worked solutions to those two questions:
7(a)
x^2 + (10  x)^2 = the area of the two shaded squares. We will try different values for x to find the minimum possible value.
f(1) = (1)^2 + (10  1)^2 = 82
f(2) = (2)^2 + (10  2)^2 = 68
f(3) = (3)^2 + (10  3)^2 = 58
f(4) = (4)^2 + (10  4)^2 = 52
f(5) = (5)^2 + (10  5)^2 = 50
f(6) = (6)^2 + (10  6)^2 = 52
f(7) = (7)^2 + (10  7)^2 = 58
f(8) = (8)^2 + (10  8)^2 = 68
f(9) = (9)^2 + (10  9)^2 = 82
Therefore, the smallest possible value of the total area of the two shaded squares = 50 square units
(You may also draw the graph of the function f(x) = x^2 + (10  x)^2 and find the minimum value of f(x). The answer will still be 50)
(b)
(before I explain the answer, you have to label the two sides that touches the line d. In my solution, I called them b and c)
In Pythagoras' theorem: a^2 = b^2 + c^2
If we apply this to the right angled triangle, then d^2 = b^2 + c^2
b^2 = area of bigger shaded square
c^2 = area of smaller shaded square
Therefore, area of bigger shaded square + area of smaller shaded square = d^2
10(a)
3 = (2)^2 + a(2) + b
3= 4 + 2a + b
2a + b = 1
4 = (5)^2 + a(5) + b
4 = 25  5a + b
5a + b = 29
(b)
(You have to solve the simultaneous equations)
2a + b = 1
 5a + b = 29
____________
7a = 28
a = 4
Sub back in to one of the equations:
2(4) + b = 1
8 + b = 1
b = 9
(c)
y = x^2 + 4x  9
y = (0)^2 + 4(0) 9
y = 9
The coordinate is (0,9)
(d)
y = x^2 + 4x  9
0 = x^2 + 4x  9
x^2 + 4x  9 = 0
ax^2 + bx + c = 0, a = 1, b = 4, c = 9
(Now use the quadratic formula and sub in the values for a, b and c)
x = (4) +/ square root((4)^2  4(1)(9)
___________________________
2(1)
x = 4 +/ square root(16 + 36)
_____________________
2
x = 4 + square root(52)
________________
2
or
x = 4  square root(52)
________________
2
x = 1.6 or x = 5.6
The two points are (1.6,0) and (5.6,0)
Hope this helps :)

RyanS
It does help. Thanks a million!! 😃

Me
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