Database- Equation of a Circle & Quadratic Equation
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This is a fun question. Perhaps you will like to do the 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.
Thank you for your kind attention.
Regards,
ahm97sic
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This question is fun. Perhaps you will like to do the question.
Worked out solution will be posted in addmaths.sgforums.com ie addmaths forum tomorrow.
Regards,
ahm97sic
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Originally posted by Ahm97sic:
This is a fun question. Perhaps you will like to do the 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.
Thank you for your kind attention.
Regards,
ahm97sic
Pt 1.Firstly, we can rearrange the equation such that it becomes:
y^2 +14y + 16xy + 13 +4x^2 + 16x= 0.And then it becomes
1(y^2) + (16x+14)(y) + (4x^2+16x+13) = 0
For it to have two distinct roots, the discriminant must > 0
(16x+14)^2-4(1)(4x^2+16x+13) > 0
(256x^2+448x+196) - (16x^2+64x+52) > 0
240x^2 + 384x + 144 > 0Since the GCD of 240, 384, 144 is 48, divide all terms by 48
5x^2 + 8x + 3 >0
Pt. 2
To find out the values of x that satisfy 5x^2 + 8x + 3 > 0,
First work out the roots
Let 5x^2 + 8x + 3 = 0.
(x+1)(5x+3) = 0
The roots are -1 and -3/5
Since 5x^2 + 8x + 3 is a U-shape graph (not an inverted-U)
Therefore, the values of x s.t. 5x^2+8x+3 are those that fall outside of the range (-1, -3/5)
x < -1 OR x > -0.6 -
Dear jayh272416,
Well done, you have solved the question.
Please come back for more interesting maths questions
ie addmaths.sgforums.com or addmaths forum.
Please come more often. OK.
Thank you for your kind attention
Regards,
ahm97sic (addmaths.sgforums.com, emaths.sgforums.com)
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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