Though the term is dreaded, as a matter of fact, quadratic equations turn up all too often in our daily real life. And, we mean, apart from the all too familiar question papers. Interestingly, the problems do not look quadratic at first glance.

To define, a quadratic equation is an equation of the form $ax^2 + bx + c = 0$, where a, b, c are real or complex numbers, x is a variable and a $\neq$ 0. If a = 0, the equation is linear, not quadratic. As it is, a quadratic equation is a polynomial whose highest power is the square of a variable x, y, etc.

Remember, not all QE are apparently quadratic in form. They can however be reduced to by simple transformation. For example, in case of $x^4 – 7x^2 + 12 = 0$, we can substitute $x^2$ with say, $t$.

### Solution & Roots of Quadratic Equations

A quadratic equation may be expressed as a product of two binomials. This gives the easy factorisation method to find solutions of a QE. How To Do: Factorize the QE and equate each factor to zero.

Let $ax^2 + bx + c = a(x – \alpha)(x – \beta) = 0.$

Then, $x = \alpha$ and $x = \beta$ will satisfy the equation.

The values of $x$ which satisfy the quadratic equation are called roots of the quadratic equation. A quadratic equation can have two and only two roots. That is, when $a \ne 0$, there are two and only two solutions to $ax^2 + bx + c = 0,$ and they are $x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$

### Proof

Let the quadratic equation be $ax^2 + bx + c = 0,$ where $a, b, c$ are real numbers and $a \neq0$.

Multiplying by $a$, we get $a^2x^2 + abx + ac = 0$

$\Rightarrow (ax)^2 + 2(ax)\frac{b}{2} + \left( \frac{b}{2} \right)^2 = \left( \frac{b}{2} \right)^2 – ac$

$\Rightarrow \left( ax + \frac{b}{2} \right)^2 = \frac{b^2 – 4ac}{4}$

Or, $\left( ax + \frac{b}{2} \right)^2 – \left( \frac{\sqrt{b^2 – 4ac}}{2} \right)^2 = 0$

$\Rightarrow \left( ax + \frac{b}{2} + \frac{\sqrt{b^2 – 4ac}}{2} \right)\left( ax + \frac{b}{2} – \frac{\sqrt{b^2 – 4ac}}{2} \right) = 0$

$\Rightarrow \left( ax – \frac{-b – \sqrt{b^2 – 4ac}}{2} \right)\left( ax – \frac{-b + \sqrt{b^2 – 4ac}}{2} \right) = 0$

Hence,

$x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$

### Nature of Roots

In a Quadratic Equation, $b^2 – 4ac$ is called discriminant and denoted by D or $\Delta$. Now for QE $ax^2 + bx + c = 0,$ where $a, b, c$ are real numbers (not necessarily rational) and $a \neq0$, the roots are:

1. Real and equal $\Leftrightarrow \Delta = 0$.
2. Real and unequal $\Leftrightarrow \Delta > 0$.
3. Imaginary $\Leftrightarrow \Delta < 0$.

When $a, b, c$ are rational numbers, the roots are:

• Real if $D \geq 0$.
• Non-real conjugate complex or imaginary numbers $\alpha \pm i\beta \Leftrightarrow D < 0,$ with $q ≠ 0.$

To further specify the first situation just above, the roots are:

1. Rational and equal $\Leftrightarrow \Delta = 0$.
2. Rational and unequal if $D > 0$ and D is also a perfect square of a rational number.
3. Unequal irrational conjugate pairs (surds) $\alpha \pm i\beta$, ($q > 0$) if $D > 0$ but not a perfect square.

### Relation Between Roots & Coefficients

Let the two roots of a quadratic equation be $\alpha$ and $\beta$ i.e., $\alpha = \frac{-b + \sqrt{b^2 – 4ac}}{2a}$ and $\beta = \frac{-b – \sqrt{b^2 – 4ac}}{2a}.$

Then, $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}.$

If $a + b + c = 0$ then $x = 1$ is one of the two roots and the other root is given by $\alpha = \frac{c}{a}.$

### Common Roots of Two Quadratic Equations

Two quadratic equations may have one or both roots common. Below is given condition for both roots being common. Let $a_1x^2 + b_1x + c_1 = 0$ and $a_2x^2 + b_2x + c_2 = 0$ be two QE having both roots common. The required condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ Two different quadratic equations cannot have only one complex or irrational root in common.

### How To Form Quadratic Equation, If Roots Are Given?

The required equation would be $x^2 – (\alpha + \beta)x + \alpha\beta = 0.$

As irrational or complex roots occur in conjugate pairs for equations with rational coefficients, the equation can be formed even when only one root is given.

## Worked Out Examples

1. Discuss the nature of the roots of $x^2 + 5x + 4 = 0.$

Solution 1:
Given: $x^2 + 5x + 4 = 0$

Here, $a = 1, b = 5, c = 4$
$\therefore \Delta = b^2 – 4ac = 9$

Thus, D = 9 is a perfect square.
Hence, the roots are real, rational and unequal.

2. Prove that for $ax^2 + by^2 = 1$ and $ax + by = 1,$ the values of $x$ will be equal if $a + b = 1.$

Solution 2:
Given, $ax + by = 1$
i.e. $y = \frac{1 – ax}{b}$

Again, $ax^2 + by^2 = 1$
$\therefore$  $ax^2 + \frac{(1 – ax)^2}{b} = 1$

$\Rightarrow ax^2(a + b) – 2ax + (1 – b) = 0$

Or, $(-2a)^2 – 4a(a + b)(1 – b) = 0$

$\Rightarrow 4a^2 – 4a(a – ab + b – b^2) = 0$

$\Rightarrow ab + b^2 = b$ (as $a \neq 0$)

Hence, $a + b = 1$ (as $b \neq 0$)

3. If the difference of the roots of the equation $x^2 + px + q = 0$ be unity, prove that $p^2 + 4q^2 = (1 + 2q)^2.$

Solution 3:

Let $\alpha$ and $\beta$ be the roots of the given eqn.
Then, $\alpha + \beta = -p$ and $\alpha\beta = q.$

Now, by the question, $\alpha – \beta = 1$

$\Rightarrow (\alpha – \beta)^2 = (1)^2$

$\Rightarrow (\alpha + \beta)^2 – 4 \alpha \beta = 1$

Or, $p^2 – 4q = 1$

$\Rightarrow p^2 = 1 + 4q$

$\Rightarrow p^2 + 4q^2 = 1 + 4q + 4q^2$

$\therefore$  $p^2 + 4q^2 = (1 + 2q)^2.$

4. In writing a quadratic equation of the form $x^2 + bx + c = 0,$ a student writes the coefficient of x incorrectly and finds the roots as 7, 8. Another student writes the constant term incorrectly and finds the roots as 8, -3. Find the correct equation.

Solution 4:
Equation in condition 1:
$x^2 -(7 + 8)x + 7 xx 8 = 0$
We get, $c = 56$

Equation in condition 2:
$x^2 – (8 – 3)x + 8 xx (-3) = 0$
We get, $b = -5$

Hence, correct eqn. is:
$x^2 – 5x + 56 = 0.$

5. Show that $ax^2 + bx + c = 0$ and $x^2 + x + 1 = 0$ cannot have a common root unless $a = b = c.$

Solution 5:
Let’s consider $x^2 + x + 1 = 0$
$\Delta = 1^2 – 4.1.1 = -3 < 0$

Therefore, the roots are non-real complex and hence, if one root is common then both the roots must be common.

Now, non-real complex roots occur in conjugate pair. As such, the two equations are not distinct.

Hence, $\frac{a}{1} = \frac{b}{1} = \frac{c}{1}$

i.e. $a = b = c$

## Multiple Choice Questions

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## Solutions

1. Given Eqn. $x^2 + px + q = 0$ and Roots: tan 30 & tan 15
So, tan 30 + tan 15 $= -p$ and tan 30.tan 15 $= q$

Now, $\tan(30 + 15) = \frac{\tan(30) + \tan(15)}{1 – \tan(30)\tan(15)}$

$\Rightarrow 1 = \frac{-p}{1 – q} \Rightarrow 1 – q = -p$
i.e.,  $1 = q – p \Rightarrow 2 + q – p = 2 + 1 = 3$

2. Let $\alpha$ and $\beta$ be the roots. Then, $\alpha + \beta = (a – 2)$ and $\alpha \beta = – (a + 1)$

Now, $\alpha^2 + \beta^2 = (\alpha + \beta)^2 – 2\alpha\beta =(a – 1)^2 + 5$
Therefore, $\alpha^2 + \beta^2$ will be minimum if $(a – 1)^2 = 0$ i.e, $a = 1$

3. Given eqn. $ix^2 – x + 12i = 0.$
Here, $a = i, b = -1, c = 12i$

$\therefore$ $x = \frac{-(-1) + \sqrt{(-1)^2 – 4(i)(12i)}}{2i}$
$\Rightarrow x = \frac{1 \pm \sqrt{1 + 48}}{2i} = \frac{1 \pm 7}{2i}$
$\Rightarrow x = \frac{4}{i}, \frac{-3}{i} = \frac{4i}{i^2}, \frac{-3i}{i^2} = -4i, 3i$

4. Eqn. is $x^2 – 7ix – 12 = 0.$

$\Rightarrow x² -3i -4i -12 = 0$
$\Rightarrow x² -3i -4i +12i² = 0$
or, $x(x-3i) -4i(x-3i) = 0$
$\Rightarrow (x-3i)(x-4i) = 0$

i.e., $x = 3i, 4i$

5. The equation is $x^2 – px + q = 0.$

Given, $p$ and $q$ are distinct prime.

Let $p=3$ and $q=2$

So the eqn. becomes
$x^2 – 3x + 2 = 0$
$\Rightarrow x^2-2x-x+2=0$
$\Rightarrow (x-1)(x-2)=0$

i.e., $x=1,2$