Concept: Quadratic and Higher Degree Equations

Example: Find the roots of the equation, x² + 12x + 20 = 0.

Solution:
Given, x² + 12x + 20 = 0
⇒ x² + 10x + 2x + 20 = 0
⇒ x(x + 10) + 2(x + 10) = 0
⇒ (x + 10)(x + 2) = 0

⇒ Either (x + 10) = 0 or (x + 2) = 0
⇒ Either x = -10 or x = -2

Hence, the roots of the equation are -10 and -2.

Example: Find the roots of the equation, 3x² – 17x + 24 = 0.

Solution:
Given, 3x² – 17x + 24 = 0
⇒ 3x² – 8x – 9x + 24 = 0
⇒ x(3x – 8) – 3(3x + 8) = 0
⇒ (3x – 8)(x – 3) = 0

⇒ x = 8/3 or 3.

Hence, the roots of the equation are 8/3 and 3.

Example: Find the roots of the equation, 2x² – 3x - 9 = 0.

Solution:
Given, 2x² – 3x - 9 = 0
⇒ 2x² – 6x + 3x - 9 = 0
⇒ 2x(x – 3) + 3(x – 3) = 0
⇒ (x – 3)(2x + 3) = 0

⇒ x = 3 or x = -3/2

Hence, the roots of the equation are -3/2 and 3.

Example: Find the integral values of x if, 3x + $\frac{5}{\mathrm{x}}$ = 8.

Solution:
3x + $\frac{5}{\mathrm{x}}$ = 8

∴ 3x² + 5 = 8x

∴3x² – 8x + 5 = 0

∴3x² - 3x – 5x + 5 = 0

∴ 3x(x – 1) – 5(x – 1) = 0

∴ (x – 1)(3x – 5) = 0

∴ x – 1 = 0 or 3x – 5 = 0

∴ x = 1 or $\frac{5}{3}$

∴ Integral value of x = 1.

Example: Find the roots of the equation, x² – 8x + 11 = 0.

Solution:
Here, splitting the middle term would not be possible, hence it is best to use the Formula method.

In the equation a = 1, b = –8, c = 11

Hence, the roots of the equation are:

x = $\frac{-\mathrm{(-8)}\pm \sqrt{{(-8)}^2-4\times 1\times 11}}{2\times 1}$

x = $\frac{8\pm \sqrt{20}}{2\times 1}$

x = $\frac{8\pm 2\sqrt{5}}{2\times 1}$

x = 4 ± √5

Hence, the two roots of the equation are (4 + √5) and (4 - √5)

Example: Find the roots of the equation, 7x² + 12x – 2 = 0.

Solution:
In the quadratic equation, 7x² + 12x – 2 = 0, we have a = 7, b = 12, c = -2.

Hence, the roots of the equation are:

x = $\frac{-12\pm \sqrt{{12}^2-4\times 7\times -2}}{2\times 7}$

x = $\frac{-12\pm \sqrt{200}}{14}$

x = $\frac{-12\pm 10\sqrt{2}}{14}$

x = $\frac{-6\pm 5\sqrt{2}}{7}$

Hence, the two roots of the equation are $\frac{-6-5\sqrt{2}}{7}$ and $\frac{-6+5\sqrt{2}}{7}$

Example:
Find the value of x if, $\frac{\mathrm{(x+1)}}{\mathrm{(x+2)}}$ + $\frac{\mathrm{(2x+1)}}{\mathrm{(x-1)}}$ = 0

Solution:
$\frac{\mathrm{(x+1)}}{\mathrm{(x+2)}}$ + $\frac{\mathrm{(2x+1)}}{\mathrm{(x-1)}}$ = 0

Multiplying both the sides by (x + 2)(x – 1),

(x + 1)(x – 1) + (2x + 1)(x + 2) = 0

∴ x² – 1 + 2x² + 4x + x + 2 = 0

∴ 3x² + 5x + 1 = 0

∴x = $\frac{-5\pm \sqrt{5^2-4\times 3\times 1}}{2\times 3}$

⇒x = $\frac{-5\pm \sqrt{13}}{6}$

Thus, the roots are $\frac{-5-\sqrt{13}}{6}$ and $\frac{-5+\sqrt{13}}{6}$

Example: There are as many oranges in a crate as there are crates in a fruit shop. Prakhar takes away one crate and then the shop has only 72 oranges left. How many oranges does each crate have?

Solution:
Let there be x oranges in each crate. Then there are x crates in the shop.

Total number of oranges in the shop = x²

Prakhar takes away one crate i.e. x oranges.

∴ x² – x = 72

∴ x² – x – 72 = 0

∴ x² – 9x + 8x – 90 = 0

∴ (x – 9)(x + 8) = 0

∴  x = 9 or x = –8

As the number of oranges cannot be negative,

∴ x = 9.

⇒ Each crate has 9 oranges.

Example: Form a quadratic equation whose roots are 4 and 5.

Solution:
A quadratic equation with roots α and β can be formed as (x - α)(x - β) = 0

∴ The given quadratic equation can be formed as (x - 4)(x - 5) = 0

⇒ x² – 4x – 5x + 20 = 0

⇒ x² – 9x + 20 = 0

Alternately,
We know a quadratic equation with roots α and β can be formed as x² – (sum of the roots)x + (product of the roots) = 0

∴ The given quadratic equation can be formed as x² – (4 + 5)x + (4 × 5) = 0

⇒ x² – 9x + 20 = 0

Example: Form a quadratic equation whose roots are 2 and -3.

Solution:
A quadratic equation with roots α and β can be formed as (x - α)(x - β) = 0

∴ The given quadratic equation can be formed as (x - 2)(x – (-3)) = 0

⇒ (x - 2)(x + 3)) = 0

⇒ x² – 2x + 3x - 6 = 0

⇒ x² + x + 6 = 0

Alternately,
We know a quadratic equation with roots α and β can be formed as x² – (sum of the roots)x + (product of the roots) = 0

∴ The given quadratic equation can be formed as x² – (2 + (-3))x + (2 × -3) = 0

⇒ x² + x - 6 = 0

Example: Form a quadratic equation whose roots are -2/3 and -1.

Solution:
The given quadratic equation is (x –(-2/3))(x-(-1)) = 0

⇒ (x+2/3)(x+1)= 0

⇒ (3x + 2)(x + 1)= 0

⇒ 3x² + 2x + 3x + 2 = 0

⇒ 3x² + 5x + 2 = 0

Alternately,
We know a quadratic equation with roots α and β can be formed as x² – (sum of the roots)x + (product of the roots) = 0

∴ The given quadratic equation can be formed as x² – (-2/3 + (-1))x + (-2/3 × -1) = 0

⇒ x² – (-5/3)x + 2/3 = 0

⇒ 3x² + 5x + 2 = 0

Example: If α and β are the roots of the quadratic equation 2x² + 3x + 1 = 0, the find the value of $\frac{1}{\mathrm{\alpha }}$ + $\frac{1}{\mathrm{\beta }}$.

Solution:
The given quadratic equation is 2x² + 3x + 1 = 0.

We know, sum of the roots is -b/a and product of the roots is c/a

∴ α + β = -3/2 and αβ = ½

Now, 

$\frac{1}{\mathrm{\alpha }}$ + $\frac{1}{\mathrm{\beta }}$ = $\frac{\mathrm{\alpha +\beta }}{\mathrm{\alpha \; \times \; \beta }}$ = $\frac{\mathrm{-3/2}}{\mathrm{1/2}}$ = -3

Alternately,
Here we can also find the roots of the equation.

Given, 2x² + 3x + 1 = 0.

⇒ 2x² + 2x + x + 1 = 0.

⇒ (2x + 1)(x + 1) = 0

⇒ x = -1/2 or -1

Let α = -1/2 and β = -1

⇒ $\frac{1}{\mathrm{\alpha }}$ + $\frac{1}{\mathrm{\beta }}$ = -2 – 1 = -3

Example: If α and β are the roots of the quadratic equation x² + 3x - 1 = 0, the find the value of $\frac{1}{{\mathrm{\alpha }}^2}$ + $\frac{1}{{\mathrm{\beta }}^2}$.

Solution:
The given quadratic equation is x² + 3x - 1 = 0.

We know, sum of the roots is -b/a and product of the roots is c/a

∴ α + β = -3 and αβ = -1

Now, 

$\frac{1}{{\mathrm{\alpha }}^2}$ + $\frac{1}{{\mathrm{\beta }}^2}$ = $\frac{{\mathrm{\alpha }}^2+{\mathrm{\beta }}^2}{{\left(\mathrm{\alpha \beta }\right)}^2}$ = $\frac{{\left(\mathrm{\alpha }+\mathrm{\beta }\right)}^2-2\mathrm{\alpha \beta }}{{\left(\mathrm{\alpha \beta }\right)}^2}$ = $\frac{9-2\times -1}{{\left(-1\right)}^2}$ = 11

Example:
If one root of the equation ax² + bx + c = 0 is triple of the other, then 3b² =
(1) 16ac            (2) c√3 a            (3) 6ac            (4) None of the above

Solution:
Given, ax² + bx + c = 0

Let the roots of the given equation are α and 3α.

∴ Sum of roots 4α = -$\frac{\mathrm{b}}{\mathrm{a}}$
⇒ α = -$\frac{\mathrm{b}}{\mathrm{4a}}$    …(1) and

Product of roots = 3α² =  $\frac{\mathrm{c}}{\mathrm{a}}$ 
⇒ 3${\left(-\frac{\mathrm{b}}{4\mathrm{a}}\right)}^2$ = $\frac{\mathrm{c}}{\mathrm{a}}$

⇒ $\frac{3{\mathrm{b}}^2}{16{\mathrm{a}}^2}$ = $\frac{\mathrm{c}}{\mathrm{a}}$

⇒ 3b² = 16ac

Hence, option 1.

Example: Find the nature of the roots of the following quadratic equations.
(1) x² + 2x - 6 = 0
(2) 3x² - 7x + 2 = 0
(3) 4x² - 12x + 9 = 0
(4) x² - 3x + 12 = 0

Solution:
The coefficients of each of these given equations are rational

(1) D = b² - 4ac = 4 – 4 × 1 × -6 = 4 + 24 = 28
Since D > 0 the roots are real and unequal.
Further, 28 is not a perfect square and hence the roots are irrational and are conjugates of each other.
Real, Distinct, and Irrational (Conjugate pairs)

(2) D = b² - 4ac = 49 – 4 × 3 × 2 = 64 – 24 = 25
Since 25 > 0, the roots are real 
Further 25 is a perfect square and hence the roots are rational.
Real, Distinct, and Rational

(3) D = b² - 4ac = 144 – 4 × 4 × 9 = 144 - 144 = 0
Since D = 0, the roots are real and equal.
Real, Equal, and Rational.

(4) D = b² - 4ac = = 9 – 4 × 1 × 12 = 9 - 48 = -37
Since D < 0, the roots are complex and are conjugates of each other.
Complex (conjugate pair)

Example: Find the nature of the roots of the quadratic equation x² - 4√3x + 12 = 0.

Solution:
The coefficients of the given equation are not rational

The roots are x = $\frac{-\left(-4\sqrt{3}\right)\pm \sqrt{{(-4\sqrt{3})}^2-4\times 1\times 12}}{2\times 1}$ = $\frac{4\sqrt{3}\pm \sqrt{48-48}}{2}$ = 2√3

Here, D = 0, hence the roots are real and equal. 

But b is irrational hence, the roots are also irrational.

Roots are Real, Equal and Irrational.

Example: f(x) is a quadratic expression such that the coefficient of x² is positive. If the roots of f(x) = 0 lie in the interval (–1, 1), then which of the following is necessarily true?
(a) f(1) > 0 & f(-1) > 0    (b) f(1) > 0 & f(-1) < 0    (c) f(1) < 0 & f(-1) > 0    (d) f(1) < 0 & f(-1) < 0

Solution:
Since the coefficient of x² is positive, the graph of the quadratic expression will be upward facing. 

Since the roots lie between -1 and 1, the graph will cut the x-axis at two points between x = -1 and x = 1.

The graph would look something like this.

​​​​​​​

We can see that f(1) and f(-1), both are greater than ‘0’.

Hence, option 1.

Example: Form a quadratic equation whose roots are 2 more than the roots of 3x² - 2x - 5 = 0.

Solution:
Here we will substitute x with x – 2.

∴ The new quadratic equation is 3(x - 2)² - 2(x - 2) - 5 = 0

⇒ 3(x² – 4x + 4) – 2x + 4 – 5 = 0

⇒ 3x² – 14x + 11 = 0

Alternately,
Given, 3x² - 2x - 5 = 0

⇒ 3x² – 5x + 3x – 5 = 0

⇒ (x + 1)(3x - 5) = 0

∴ The roots of the given equation are -1 and 5/3

⇒ Roots of the new quadratic equation are -1 + 2 = 1 and 5/3 + 2 = 11/3

∴ The new equation is (x - 1)(x – 11/3) = 0

⇒ (x - 1)(3x - 11) = 0

⇒ 3x² – 14x + 11 = 0

Example: Form a quadratic equation whose roots are 2 less than the roots of 3x² - 2x - 5 = 0.

Solution:
Here we will substitute x with x + 2.

∴ The new quadratic equation is 3(x + 2)² - 2(x + 2) - 5 = 0

⇒ 3(x² + 4x + 4) – 2x - 4 – 5 = 0

⇒ 3x² + 10x + 3 = 0

Alternately,
Given, 3x²-2x-5=0

⇒ 3x² – 5x + 3x – 5 = 0

⇒ (x + 1)(3x - 5) = 0

∴ The roots of the given equation are -1 and 5/3

⇒ Roots of the new quadratic equation are -1 - 2 = -3 and 5/3 - 2 = -1/3

∴ The new equation is (x – (-3))(x – (-1/3)) = 0

⇒ (x + 3)(3x + 1) = 0

⇒ 3x² + 10x + 3 = 0

Example: Form a quadratic equation whose roots are three times the roots of 3x²-2x-5=0.

Solution:
Here we will substitute x with x/3.

∴ The new quadratic equation is 3(x/3)² - 2(x/3) - 5 = 0

⇒ x²/3 - 2x/3 - 5 = 0

⇒ x² - 2x - 15 = 0

Alternately,
Given, 3x²-2x-5=0

⇒ 3x² – 5x + 3x – 5 = 0

⇒ (x + 1)(3x - 5) = 0

∴ The roots of the given equation are -1 and 5/3

⇒ Roots of the new quadratic equation are -1 × 3 = -3 and 5/3 × 3 = 5

∴ The new equation is (x – (-3))(x – 5) = 0

⇒ (x + 3)(x – 5) = 0

⇒ x² - 2x - 15 = 0

Example: Form a quadratic equation whose roots are half the roots of 3x²-2x-5=0.

Solution:
Here we will substitute x with 2x.

∴ The new quadratic equation is 3(2x)² - 2(2x) - 5 = 0

⇒ 12x² - 4x - 5 = 0

Alternately,
Given, 3x²-2x-5=0

⇒ 3x² – 5x + 3x – 5 = 0

⇒ (x + 1)(3x - 5) = 0

∴ The roots of the given equation are -1 and 5/3

⇒ Roots of the new quadratic equation are -1 ÷ 2 = -1/2 and 5/3 ÷ 2 = 5/6

∴ The new equation is (x – (-1/2))(x – 5/6) = 0

⇒ (2x + 1)(6x - 5) = 0

⇒ 12x² - 4x - 5 = 0

Example: The number of roots common between the two equations x³ - 6x² + 11x - 6 = 0 and x3 + 2x² - x - 2 = 0 is:
(1) 0            (2) 1            (3) 2             (4) 3

Solution:
f(x) = x³ - 6x² + 11x - 6 = 0   ...(1)
g(x) = x³ + 2x² - x - 2 = 0   ...(2)

Equating equations f(x) and g(x), we get,

⇒ x³ - 6x² + 11x - 6 = x3 + 2x² - x - 2

⇒ 8x² – 12x + 4 = 0

⇒ 2x² – 3x + 1 = 0

⇒ (2x – 1)(x – 1) = 0

∴ x = 1/2 or x = 1

Only for x = 1, both f(x) = 0 and g(x) = 0

Hence, option 2.

Example: The number of roots common between the two equations x3 + 3x² + 4x + 5 = 0 and x3 + 2x² + 7x + 3 = 0 is:
(1) 0            (2) 1            (3) 2             (4) 3

Solution:
f(x) = x3 + 3x² + 4x + 5 = 0   ...(1)
g(x) = x3 + 2x² + 7x + 3 = 0   ...(2)

Step 1: Equating equations f(x) and g(x), we get,

⇒ x3 + 3x² + 4x + 5 = x3 + 2x² + 7x + 3

⇒ x² – 3x + 2 = 0

⇒ (x – 1)(x – 2) = 0

∴ x = 1 or x = 2

Since both x = 1 and x = 2, do not satisfy either (1) or (2), there exists no common roots for equations (1) and (2).

Hence, option 1.

Example: Find the value(s) of ‘m’, so that the equations x² – x – 12 = 0 & mx² + 10x + 3 = 0 have exactly one root in common.
(a) 3                      (b) -43/16        (c) 4                    (d) Either (a) or (b)

Solution:
f(x) = x² – x – 12
g(x) = mx² + 10x + 3

Solving for f(x) 
⇒ (x + 3)(x - 4) = 0

⇒ x = -3 or 4

The common root will be either -3 or 4

Case 1: -3 is the common root

⇒ g(x) = mx² + 10x + 3 = 0

⇒ 9m – 30 + 3 = 0

⇒ m = 3

Case 2: 4 is the common root

⇒ g(x) = mx² + 10x + 3 = 0

⇒ 16m + 40 + 3 = 0

⇒ m = -43/16

Hence, option 4.