CIE AS MATHS QUADRATIC QUESTIONS

[xxemoxx]

If curve y=p-(x-q)^2 crosses x-axis at (-1,0) and (5,0). Find maximum values of y.

another question

What values of m makes the following expression in perfect square.
(2m-1)x^2 - 4(m-2)x + (m-1)

[vanibharutham]

For the first one,

first thing u need to do is find the values of p and q, although p is not really necessary, but we shall find them anyways

substitute in the values for x and y into

y = p - (x - q)²

At (-1,0)
0 = p - ( -1 - q )²
0 = p - (q² + 2q + 1)
p = q² + 2q + 1

At (5,0)
0 = p - ( 5 - q )²
0 = p - (25 - 10q +q²)
p = q² - 10q + 25

Equate both the p's

q² - 10q + 25 = q² + 2q + 1
Giving,

- 12q = -24
q = 2

q² + 2q + 1 = p
(2)²+2(2) + 1 = p
4 + 4 + 1 = p
p = 9

now we know that

y =  9 - (x - 2)²

to find the maximum values, you need to differentiate

dy/dx = [USING CHAIN RULE]
======> -2(x-2)(1)
=> -2(x-2)

at maximum values, dy/dx = 0

therefore

-2(x-2) = 0
x- 2 = 0
x = 2

y = 9 - (2 - 2)²
y = 9

therefore the coordinates of maximum are (2,9)

[vanibharutham]

And the second one,

for any equation to be a "perfect square" there needs to be only one root

therefore,

b² = 4ac

our equation, in its simplified form

(2m-1)x² + (-4m +8) x + (m-1)

b = (-4m+8)
a = (2m-1)
c = (m-1)

b² = 4ac
(-4m+8)² = 4(2m-1)(m-1)
16m² - 64m + 64 = 4(2m²-3m + 1)
4m² - 16m + 16 = 2m² - 3m + 1
2m² - 13m + 15 = 0
(2m - 3)(m - 5) = 0

Therefore,

the values at which m is a perfect square is when m = 1½ and when m = 5