4.1-Quadratic Equations

4.1-Quadratic Equations Important Formulae

You are currently studying
Grade 10 → Math → Quadratic Equations → 4.1-Quadratic Equations

After successful completion of this topic, you should be able to:

  • Write Quadratic Equation in order to represent the given situation algebraically.
  • Rewrite the given equations in the standard form in order to check whether they are quadratic or not.

A quadratic equation is a polynomial equation of degree 2, which can be expressed in the standard form:

$ax^2 + bx + c = 0$

where:

  • $a$, $b$, and $c$ are constants
  • $a \neq 0$ (if $a = 0$, the equation becomes linear)
  • $x$ represents the variable

The solutions of a quadratic equation are known as the roots of the equation. These roots can be real or complex, depending on the value of the discriminant, which is given by:

$D = b^2 - 4ac$

The nature of the roots can be determined as follows:

  • If $D > 0$, the equation has two distinct real roots.
  • If $D = 0$, the equation has exactly one real root (also known as a repeated or double root).
  • If $D < 0$, the equation has two complex roots.

Quadratic equations can be classified into different forms:

  • Standard Form: $ax^2 + bx + c = 0$
  • Vertex Form: $y = a(x - h)^2 + k$, where $(h, k)$ is the vertex of the parabola.
  • Factored Form: $y = a(x - r_1)(x - r_2)$, where $r_1$ and $r_2$ are the roots of the equation.

The graphical representation of a quadratic equation is a parabola. The general characteristics of a parabola include:

  • Direction: If $a > 0$, the parabola opens upwards; if $a < 0$, it opens downwards.
  • Vertex: The highest or lowest point of the parabola, depending on its direction.
  • Axis of Symmetry: The vertical line that passes through the vertex, given by $x = -\frac{b}{2a}$.

To find the roots of a quadratic equation, several methods can be employed:

  • Factorisation: This involves expressing the quadratic in the form $(x - r_1)(x - r_2) = 0$.
  • Completing the Square: This method rewrites the equation in the vertex form.
  • Quadratic Formula: The roots can be directly calculated using the formula:

    • $x = \frac{-b \pm \sqrt{D}}{2a}$

Examples of quadratic equations include:

  • $x^2 - 5x + 6 = 0$
  • $2x^2 + 3x - 2 = 0$
  • $-x^2 + 4x + 8 = 0$

Quadratic equations are significant in various fields such as physics, engineering, and economics, as they model a wide range of phenomena including projectile motion, area calculations, and optimization problems. Understanding their properties and methods of solutions is essential for further studies in mathematics and its applications.


Chompinha, CC BY-SA 4.0, via Wikimedia Commons

Check whether the following are quadratic equations:

(i) $(x+1)^2 =2(x–3)$
(ii) $x^2 –2x=(–2)(3–x)$
(iii) $(x–2)(x+1)=(x–1)(x+3)$
(iv) $(x–3)(2x+1)=x(x+5) $
(v) $(2x–1)(x–3)=(x+5)(x–1) $
(vi) $x^2 +3x+1=(x–2)^2$
(vii) $(x+2)^3 =2x(x^2 –1)$
(viii) $x^3 –4x^2 –x+1=(x–2)^3$

Solution:

Check whether the following are quadratic equations:


(i) $(x+1)^2 = 2(x-3)$
Expand both sides:
$(x+1)^2 = x^2 + 2x + 1$
$2(x-3) = 2x - 6$
Thus, the equation becomes:
$x^2 + 2x + 1 = 2x - 6$
Simplifying:
$x^2 + 2x + 1 - 2x + 6 = 0$
$x^2 + 7 = 0$
This is a quadratic equation.

(ii) $x^2 - 2x = (-2)(3-x)$
Expand both sides:
$(-2)(3-x) = -6 + 2x$
Thus, the equation becomes:
$x^2 - 2x = -6 + 2x$
Simplifying:
$x^2 - 2x - 2x + 6 = 0$
$x^2 - 4x + 6 = 0$
This is a quadratic equation.

(iii) $(x-2)(x+1) = (x-1)(x+3)$
Expand both sides:
$(x-2)(x+1) = x^2 + x - 2x - 2 = x^2 - x - 2$
$(x-1)(x+3) = x^2 + 3x - x - 3 = x^2 + 2x - 3$
Thus, the equation becomes:
$x^2 - x - 2 = x^2 + 2x - 3$
Simplifying:
$x^2 - x - 2 - x^2 - 2x + 3 = 0$
$-3x + 1 = 0$
This is a linear equation, not quadratic.

(iv) $(x-3)(2x+1) = x(x+5)$
Expand both sides:
$(x-3)(2x+1) = 2x^2 + x - 6x - 3 = 2x^2 - 5x - 3$
$x(x+5) = x^2 + 5x$
Thus, the equation becomes:
$2x^2 - 5x - 3 = x^2 + 5x$
Simplifying:
$2x^2 - 5x - 3 - x^2 - 5x = 0$
$x^2 - 10x - 3 = 0$
This is a quadratic equation.

(v) $(2x-1)(x-3) = (x+5)(x-1)$
Expand both sides:
$(2x-1)(x-3) = 2x^2 - 6x - x + 3 = 2x^2 - 7x + 3$
$(x+5)(x-1) = x^2 - x + 5x - 5 = x^2 + 4x - 5$
Thus, the equation becomes:
$2x^2 - 7x + 3 = x^2 + 4x - 5$
Simplifying:
$2x^2 - 7x + 3 - x^2 - 4x + 5 = 0$
$x^2 - 11x + 8 = 0$
This is a quadratic equation.

(vi) $x^2 + 3x + 1 = (x-2)^2$
Expand both sides:
$(x-2)^2 = x^2 - 4x + 4$
Thus, the equation becomes:
$x^2 + 3x + 1 = x^2 - 4x + 4$
Simplifying:
$x^2 + 3x + 1 - x^2 + 4x - 4 = 0$
$7x - 3 = 0$
This is a linear equation, not quadratic.

(vii) $(x+2)^3 = 2x(x^2 - 1)$
Expand both sides:
$(x+2)^3 = x^3 + 6x^2 + 12x + 8$
$2x(x^2 - 1) = 2x^3 - 2x$
Thus, the equation becomes:
$x^3 + 6x^2 + 12x + 8 = 2x^3 - 2x$
Simplifying:
$x^3 + 6x^2 + 12x + 8 - 2x^3 + 2x = 0$
$-x^3 + 6x^2 + 14x + 8 = 0$
This is a cubic equation, not quadratic.

(viii) $x^3 - 4x^2 - x + 1 = (x-2)^3$
Expand both sides:
$(x-2)^3 = x^3 - 6x^2 + 12x - 8$
Thus, the equation becomes:
$x^3 - 4x^2 - x + 1 = x^3 - 6x^2 + 12x - 8$
Simplifying:
$x^3 - 4x^2 - x + 1 - x^3 + 6x^2 - 12x + 8 = 0$
$2x^2 - 13x + 9 = 0$
This is a quadratic equation.

Suggest improvements/Report errors

Represent the following situations in the form of quadratic equations:

(i) The area of a rectangular plot is 528 m2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

Solution:

(i) The area of a rectangular plot is 528 m2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
Let the breadth of the plot be $x$ metres. Then the length of the plot is $(2x + 1)$ metres. The area of the rectangle is given by: $$ \text{Area} = \text{Length} \times \text{Breadth} $$ Substituting the given values: $$ 528 = (2x + 1) \times x $$ Simplifying: $$ 528 = 2x^2 + x $$ Rearranging the equation: $$ 2x^2 + x - 528 = 0 $$ ---
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
Let the integers be $x$ and $(x + 1)$. The product of these integers is given by: $$ x(x + 1) = 306 $$ Simplifying: $$ x^2 + x - 306 = 0 $$ ---
(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
Let Rohan's present age be $x$ years. Then his mother's present age is $(x + 26)$ years. In 3 years, Rohan's age will be $(x + 3)$ and his mother's age will be $(x + 26 + 3) = (x + 29)$. The product of their ages in 3 years is given by: $$ (x + 3)(x + 29) = 360 $$ Expanding: $$ x^2 + 29x + 3x + 87 = 360 $$ Simplifying: $$ x^2 + 32x - 273 = 0 $$ ---
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Let the speed of the train be $x$ km/h. The time taken to cover 480 km at speed $x$ is: $$ \frac{480}{x} \text{ hours} $$ If the speed had been 8 km/h less, the speed would be $(x - 8)$ km/h, and the time taken would be: $$ \frac{480}{x - 8} \text{ hours} $$ According to the problem, the time taken with the reduced speed is 3 hours more, so: $$ \frac{480}{x - 8} - \frac{480}{x} = 3 $$ Simplifying: $$ \frac{480x - 480(x - 8)}{x(x - 8)} = 3 $$ Expanding the numerator: $$ \frac{480x - 480x + 3840}{x(x - 8)} = 3 $$ Simplifying further: $$ \frac{3840}{x(x - 8)} = 3 $$ Multiplying both sides by $x(x - 8)$: $$ 3840 = 3x(x - 8) $$ Expanding: $$ 3840 = 3x^2 - 24x $$ Rearranging the equation: $$ 3x^2 - 24x - 3840 = 0 $$ Dividing the entire equation by 3: $$ x^2 - 8x - 1280 = 0 $$

Suggest improvements/Report errors