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 \(ax^2 +bx + c =0\) where \(a\neq 0\), are solved using the quadratic formula, \(x = {b \pm \sqrt{b^24ac} \over 2a}\) 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, \(b^24ac\), 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 \(\begin{align} x = \frac{b \pm \sqrt{b^24ac}}{2a} \end{align}\)
\(\begin{align} b^2 – 4ac \end{align}\) 
Roots 
Nature Of Roots 

\(\gt0\) 
\(\begin{align} x = {b + \sqrt{b^24ac} \over 2a} \end{align}\) \(\begin{align} x = {b  \sqrt{b^24ac} \over 2a} \end{align}\) 
2 real & distinct roots 
\(=0\) 
\(\begin{align} x &= \frac{b \pm \sqrt{0}} {2a} \\ &=\frac{b}{2a} \end{align}\) 
2 real & repeated root 
\(\lt0\) 
\(\begin{align} x = {b \pm \sqrt{\text{ve}} \over 2a} \end{align}\) 
Roots are

Solving Linear Simultaneous Equations
Question 1:
Solve these simultaneous equations.
\(\begin{align} y &= 2x1 \\ y &= x+5 \end{align}\)
Solution:
You can solve this using the Substitution Method.
\(\begin{align} y &= 2x1 &  \;(1) \\ y &= x+5 & \;(2) \end{align}\)
Since, both equations equate to \(y\) hence you can substitute the value of (1) in (2)
\(\begin{align} 2x  1 &= x+5\\ \implies\; 2x + x &= 1 + 5\\ \implies\qquad 3x &= 6 \\ \implies\qquad\;\; x &=2 \;\;\;\text{ is the root} \end{align}\)
Solving Linear & Non–Linear Simultaneous Equation
Question 2:
Solve these simultaneous equations
\(\begin{align} y &= x^2 +2x 5 \\ y &= x+5 \end{align}\)
Solution:
You can solve this using the Substitution Method.
\(\begin{align} y &= x^2 +2x 5 & \;(1)\\ y &= x+5 & \; (2) \end{align}\)
Since both equations equate to \(y\) hence you can substitute the value of (1) in (2)
\(\begin{align} x+5 &= x^2+2x5\\ \implies x^2 +2x 5 + x 5 &= 0\\ \implies\qquad\quad x^2 +3x 10 &= 0\\ \\ \end{align}\)
\(\begin{align} \therefore\quad x &= {(3) \pm \sqrt{3^24(1)(10)} \over 2(1)} \\ \implies x &= {3 \pm \sqrt{49} \over 2} \\ \implies x &=2 \qquad \text{or} \qquad 5 \end{align}\)
Thus, the discriminant \(b^2 4ac = 49\) which is \(> 0\) and hence there are two real roots, \(2\) and \(–5\).
Question 3:
Solve these simultaneous equations
\(\begin{align} y &= x^2 11x +30 \\ y &= x +5 \end{align}\)
Solution:
You can solve this using the Substitution Method.
\(\begin{align} y &= x^2 11x +30 & \;(1)\\ y &= x +5 & \;(2) \end{align}\)
Since both equations equate to \(y\) hence you can substitute the value of (1) in (2)
\(\begin{align} x^211x+30 &= x+5\\ \implies\;\; x^2 11x +30 +x5 &= 0\\ \implies\qquad\qquad x^2 10x +25 &= 0 \end{align}\)
\(\begin{align} \therefore\quad x &= {(10) \pm \sqrt{(10)^24(1)(25)} \over 2(1)} \\ \implies x &= {10 \pm \sqrt{0} \over 2}\\ \implies x &=5 \end{align}\)
Thus, the discriminant is \(b^2 4ac = 0\) which is \( = 0\) and hence there is one real distinct root, 5.
Question 4:
Solve these simultaneous equations
\(\begin{align} y &= x^2 x +6 \\ y &= x +5 \end{align}\)
Solution:
You can solve this using the Substitution Method.
\(\begin{align} y &= x^2 x +6 & \;(1)\\ y &= x +5 & \;(2) \end{align}\)
Since both equations equate to \(y\) hence you can substitute the value of (1) in (2).
\(\begin{align} x^2x+6 &= x+5\\ \implies x^2 x +6 +x5 &= 0\\ \implies\qquad\quad x^2 +0x+1 &= 0 \end{align}\)
\(\begin{align} \therefore\quad\qquad x &= {(0) \pm \sqrt{(0)^24(1)(1)} \over 2(1)}\\ \implies\qquad x &= {(0) + \sqrt{4} \over 2} \\ \implies\qquad x &= {(0)  \sqrt{4} \over 2} \end{align}\)
As the discriminant \(b^2 4ac = 4\) 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
\(\begin{align} y &=px^2 +qx +r \\ y &= ax +b \end{align}\)
You can solve them using \(\begin{align} x = {b \pm \sqrt{b^24ac} \over 2a} \end{align}\)
Example 1:
Find the value of \(m\) for which the line \(y = mx 3 \) is a tangent to the curve \(\begin{align} y=x+\frac {1}{x} \end{align}\).
Solution:
When we look at the question, it mentions a tangent to the curve \(y=x+\frac {1}{x}\) which means the curve gets cut only at one point. It means that \(b^2 4ac = 0\) and there is only one real root.
\(\begin{align} y &= mx 3 & \;(1)\\ y &=x+\frac {1}{x} & \;(2) \end{align}\)
\(\begin{align} \implies\qquad\qquad\quad\; mx3 &= x+ \frac{1}{x}\\ \implies\qquad\qquad\quad\; mx 3 &= \frac{x^2+1}{x}\\ \implies\qquad\qquad\; mx^23x &= x^2 +1\\ \implies\; x^2+1 mx^2 +3x &=0\\ \implies (1m)x^2 +3x +1 &=0 \end{align}\)
Now we have \(a=(1m),\; b=3,\; c=1\)
We already know \(b^2 4ac = 0\)
Hence,
\(\begin{align} \implies 3^2 4(1m)(1)&=0\\ \implies\qquad\qquad\;\;\, 1m &=\frac{9}{4} \\ \implies\qquad\qquad\qquad\;\, m &=1\frac{9}{4} \\ \implies\qquad\qquad\qquad\;\, m &=\frac{5}{4} \end{align}\)
Question 5:
Find the value of \(m\) for which the line \(y = x+m\) is a tangent to the curve \(y = 2x^2 1\).
Solution:
When we look at the question, it mentions a tangent to the curve \(y = 2x^2 1\), which means the curve gets cut only at one point. It means that \(b^2 4ac = 0\) and there is only one real root.
Equating both equations with each other we get,
\(\begin{align} x+m &=2x^2 1\\ \implies\;\;\, 2x^2 xm1 &= 0\\ \implies 2x^2 x(m+1 ) &= 0 \end{align}\)
Hence, we obtain \(a=2,\; b=1 \;\text{and} \; c=(m+1)\)
\(\begin{align} \therefore\qquad(1)^2 4(2)(m1)&=0\\ \implies\qquad\qquad 18(m1) &= 0\\ \implies\qquad\qquad\quad\;\; 1+8m +8 &= 0\\ \implies\qquad\qquad\qquad\quad\; 8m +9 &= 0\\ \implies\qquad\qquad\qquad\qquad\quad\;\; m &= \frac{9}{8}\\ \implies\qquad\qquad\qquad\qquad\quad\;\ m &= 1\frac{1}{8} \end{align}\)
Conclusion
In this article, we have observed the relationship between the number of points of intersection and 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!'
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