Equations And Inequalities
In this article, we are going to learn about Equalities and Inequalities:
- Relationship between the number of points of intersection and the nature of solutions of a pair of simultaneous equations
- Related conditions for a given line to
- intersect a given curve
- be a tangent to a given curve
- not intersect a given curve
This article is specifically written to meet the requirements of Secondary 3 Additional Mathematics.
Quadratic Formula
The general solution of the equation ax2+bx+c=0 where a≠0, are solved using the quadratic formula, x=−b±√b2−4ac2a where x is called the roots of the equation. We will be using this to understand the nature of the roots.
Discriminant and Nature of Roots
A discriminant, b2−4ac, is a part of the quadratic formula which can be used to tell us the number of roots a quadratic equation has and nature refers to the type of roots that exist.
For the quadratic formula x=−b±√b2−4ac2a
b2–4ac |
Roots |
Nature Of Roots |
---|---|---|
>0 |
x=−b+√b2−4ac2a x=−b−√b2−4ac2a |
2 real & distinct roots |
=0 |
x=−b±√02a=−b2a |
2 real & repeated root |
<0 |
x=−b±√-ve2a |
Roots are
|
Solving Linear Simultaneous Equations
Question 1:
Solve these simultaneous equations.
y=2x−1y=−x+5
Solution:
You can solve this using the Substitution Method.
y=2x−1−−−−(1)y=−x+5−−−−(2)
Since both equations equate to y hence you can substitute the value of (1) in (2)
2x−1=−x+5⟹2x+x=1+5⟹3x=6⟹x=2 is the root
Solving Linear & Non–Linear Simultaneous Equation
Question 2:
Solve these simultaneous equations
y=x2+2x−5y=−x+5
Solution:
You can solve this using the Substitution Method.
y=x2+2x−5−−−−(1)y=−x+5−−−−(2)
Since both equations equate to y hence you can substitute the value of (1) in (2)
−x+5=x2+2x−5x2+2x−5+x−5=0x2+3x−10=0
x=−(3)±√32−4(1)(−10)2(1)x=−3±√492x=2or−5
Thus, the discriminant b2−4ac=49 which is >0 and hence there are two real roots, 2 and –5.
Question 3:
Solve these simultaneous equations
y=x2−11x+30y=−x+5
Solution:
You can solve this using the Substitution Method.
y=x2−11x+30−−−−(1)y=−x+5−−−−(2)
Since both equations equate to y hence you can substitute the value of (1) in (2)
x2−11x+30=−x+5x2−11x+30+x−5=0x2−10x+25=0
x=−(−10)±√(−10)2−4(1)(25)2(1)x=10±√02x=5
Thus, the discriminant is b2−4ac=0 which is =0 and hence there is one real distinct root, 5.
Question 4:
Solve these simultaneous equations
y=x2−x+6y=−x+5
Solution:
You can solve this using the Substitution Method.
y=x2−x+6−−−−(1)y=−x+5−−−−(2)
Since both equations equate to y hence you can substitute the value of (1) in (2).
x2−x+6=−x+5x2−x+6+x−5=0x2+0x+1=0
x=−(0)±√(0)2−4(1)(1)2(1)x=−(0)+√−42x=−(0)−√−42
As the discriminant b2−4ac=−4 which is <0 and hence there are no real distinct roots or no solution possible.
Related conditions for determining the number of Points of Intersection of a Line and a Curve
When two equations are given in the form
y=px2+qx+ry=ax+b
You can solve them using x=−b±√b2−4ac2a
Example 1:
Find the value of m for which the line y=mx−3 is a tangent to the curve y=x+1x.
Solution:
When we look at the question, it mentions a tangent to the curve y=x+1x which means the curve gets cut only at one point. It means that b2−4ac=0 and there is only one real root.
y=mx−3−−−−(1)y=x+1x−−−−(2)
mx−3=x+1xmx−3=x2+1xmx2−3x=x2+1x2+1−mx2+3x=0(1−m)x2+3x+1=0
Now we have a=(1−m),b=3,c=1
We already know b2−4ac=0
Hence,
32−4(1−m)(1)=0(1−m)=94m=1−94m=−54
Question 5:
Find the value of m for which the line y=x+m is a tangent to the curve y=2x2−1.
Solution:
When we look at the question, it mentions a tangent to the curve y=2x2−1, which means the curve gets cut only at one point. It means that b2−4ac=0 and there is only one real root.
Equating both equations with each other we get,
x+m=2x2−12x2−x−m−1=02x2−x−(m+1)=0
Hence, we obtain a=2,b=−1andc=−(m+1)
(−1)2−4(2)(−m−1)=01−8(−m−1)=01+8m+8=08m+9=0m=−98m=−118
Conclusion
In this article, we have observed the relationship between the number of points of intersection and the nature of solutions of a pair of simultaneous equations as per the Secondary 3 Additional Mathematics syllabus in Singapore.
We have also studied the related conditions for a given line to
- intersect a given curve
- be a tangent to a given curve
- not intersect a given curve
Multiple examples and questions are also given to aid in understanding these concepts better. Keep learning! Keep improving!
Continue Learning | |
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Quadratic Functions in Real-World Context | Equations and Inequalities |
Logarithmic Functions | Surds |
Polynomials & Cubic Equations | Partial Fraction |
Exponential Functions | Coordinate Geometry (Circles) |
Linear Law | Binomial Theorem |