# Equations 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

The general solution of the equation $ax^2 +bx + c =0$  where $a\neq 0$, are solved using the quadratic formula, $x = {-b \pm \sqrt{b^2-4ac} \over 2a}$  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, $b^2-4ac$, 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 \begin{align} x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \end{align}

\begin{align} b^2 – 4ac \end{align}

Roots

Nature Of Roots

$\gt0$

\begin{align} x = {-b + \sqrt{b^2-4ac} \over 2a} \end{align}

\begin{align} x = {-b - \sqrt{b^2-4ac} \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

• Imaginary
• Complex
• No real roots

## Solving Linear Simultaneous Equations

Question 1:

Solve these simultaneous equations.

\begin{align} y &= 2x-1 \\ y &= -x+5 \end{align}

Solution:

You can solve this using the Substitution Method.

\begin{align} y &= 2x-1 & ---- \;(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+2x-5\\ x^2 +2x -5 + x -5 &= 0\\ x^2 +3x -10 &= 0\\ \\ \end{align}

\begin{align} x &= {-(3) \pm \sqrt{3^2-4(1)(-10)} \over 2(1)} \\ x &= {-3 \pm \sqrt{49} \over 2} \\ 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^2-11x+30 &= -x+5\\ x^2 -11x +30 +x-5 &= 0\\ x^2 -10x +25 &= 0 \end{align}

\begin{align}​​ x &= {-(-10) \pm \sqrt{(-10)^2-4(1)(25)} \over 2(1)} \\ x &= {10 \pm \sqrt{0} \over 2}\\ 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^2-x+6 &= -x+5\\ x^2 -x +6 +x-5 &= 0\\ x^2 +0x+1 &= 0 \end{align}

\begin{align}​​ x &= {-(0) \pm \sqrt{(0)^2-4(1)(1)} \over 2(1)}\\ x &= {-(0) + \sqrt{-4} \over 2} \\ 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^2-4ac} \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} mx-3 &= x+ \frac{1}{x}\\ mx -3 &= \frac{x^2+1}{x}\\ mx^2-3x &= x^2 +1\\ x^2+1 -mx^2 +3x &=0\\ (1-m)x^2 +3x +1 &=0 \end{align}

Now we have  $a=(1-m),\; b=3,\; c=1$

We already know $b^2 -4ac = 0$

Hence,

\begin{align}​​ 3^2 -4(1-m)(1)&=0\\ (1-m) &=\frac{9}{4} \\ m &=1-\frac{9}{4} \\ 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\\ 2x^2 -x-m-1 &= 0\\ 2x^2 -x-(m+1 ) &= 0 \end{align}

Hence, we obtain $a=2,\; b=-1 \;\text{and} \; c=-(m+1)$

\begin{align} (-1)^2 -4(2)(-m-1)&=0\\ 1-8(-m-1) &= 0\\ 1+8m +8 &= 0\\ 8m +9 &= 0\\ m &= -\frac{9}{8}\\ m &= -1\frac{1}{8} \end{align}

## 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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