This is the worked solution for
the fun question.
Question
The equation of
a circle is
4x^2 + 16xy +
y^2 +16x + 14y + 13 = 0.
If a real value
of x is substituted, the equation of the circle becomes a quadratic
equation in y.
Given that the
quadratic equation in y has two distinct roots, show
that
5x^2 + 8x + 3
> 0
and find the
range of the values of x which satisfy this inequality.
Answer
Step 1 :
Since the question asks
for a quadratic equation in y, we will re-arrange it into the
format of a quadratic equation in y ie ay^2 + by + c =
0
4x^2 + 16xy +
y^2 +16x + 14y + 13 = 0.
y^2 +14y +
16xy + 13 +4x^2 + 16x= 0
1(y^2) + (16x+14)(y) + (4x^2+16x+13) = 0 ie ay^2 + by
+ c = 0
Step 2 :
Since the question says
that the quadratic equation in y has two distinct roots, so the
discriminant b^2 - 4ac > 0
Since
a = 1, b = 16x + 14, c = 4x^2+16x+13,
b^2 - 4ac >
0
(16x + 14)^2 -
4(1)(4x^2+16x+13) > 0
256x^2 + 448x +
196 - 16x^2 - 64x - 52 > 0
240x^2 + 384x +
144 > 0
Divide by 48
throughout
5x^2 + 8x + 3
>0 (Shown)
Step 3 :
To find the range of the
values of x which satisfy this inequality
5x^2 + 8x + 3
> 0
(x+1)(5x+3) >
0
There are three
main approaches to solve it. One approach is to use the number line
approach, two is to sketch the curve and three is the short-cut
method. The short-cut-method will be used.
Short-cut
Method
(1)
(x+a)(x-b) > 0
x
< - a or x > b
(2)
(x+a)(x-b) < 0
-a < x < b
Since (x+1)(5x+3) >
0
x < -1 or x > - 3/5
Thank you for
your kind attention.
Regards,
ahm97sic
