**Quadratic Equation**

Equations that are of order 2 are referred to as **Quadratic Equation**. Here on this page, we have provided the formulas as well as tricks for solving the Quadratic Equation in a step by step way. Individuals can also practice from the given Quadratic Equation Algebra Questions to gain good marks in the examination.

**Quadratic Equation**

**Quadratic Equation Definition:**

Quadratic equations are the polynomial equations of degree 2 in one variable of type f(x) = ax^{2} + bx + c where a, b, c, ∈ R and a ≠ 0.

Where

- ‘a’ is called the leading coefficient
- ‘c’ is called the absolute term of f (x).
- The values of x satisfying the quadratic equation are the roots of the quadratic equation (α,β).

The quadratic equation will always have two roots. The nature of roots may be either real or imaginary.

** General Form:** ax

^{2}+ bx + c = 0

__Examples__: 3x^{2} + x + 5 = 0, -x^{2} + 7x + 5 = 0, x^{2} + x = 0.

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**Quadratic Equation Formula:**

The solution or roots of a quadratic equation are given by the quadratic formula:

(α, β) = [-b ± √(b^{2} – 4ac)]/2ac

**Formulas for Solving Quadratic Equations:**

- The roots of the quadratic equation: x = (-b ± √D)/2a, where D = b
^{2}– 4ac - Nature of roots:

- D > 0, roots are real and distinct (unequal)
- D = 0, roots are real and equal (coincident)
- D < 0, roots are imaginary and unequal

- The roots (α + iβ), (α – iβ) are the conjugate pair of each other.
- Sum and Product of roots: If α and β are the roots of a quadratic equation, then

- S = α+β= -b/a = coefficient of x/coefficient of x
^{2} - P = αβ = c/a = constant term/coefficient of x
^{2}

- Quadratic equation in the form of roots: x
^{2}– (α+β)x + (αβ) = 0 - The quadratic equations a
_{1}x_{2}+ b_{1}x + c_{1}= 0 and a_{2}x_{2}+ b_{2}x + c_{2}= 0 have;

- One common root if (b
_{1}c_{2}– b_{2}c_{1})/(c_{1}a_{2}– c_{2}a_{1}) = (c_{1}a_{2}– c_{2}a_{1})/(a_{1}b_{2}– a_{2}b_{1}) - Both roots common if a
_{1}/a_{2}= b_{1}/b_{2}= c_{1}/c_{2}

- In quadratic equation ax
^{2 }+ bx + c = 0 or [(x + b/2a)^{2}– D/4a^{2}]

- If a > 0, minimum value = 4ac – b
^{2}/4a at x = -b/2a. - If a < 0, maximum value 4ac – b
^{2}/4a at x= -b/2a.

- If α, β, γ are roots of cubic equation ax
^{3}+ bx^{2}+ cx + d = 0, then, α + β + γ = -b/a, αβ + βγ + λα = c/a, and αβγ = -d/a - A quadratic equation becomes an identity (a, b, c = 0) if the equation is satisfied by more than two numbers i.e. having more than two roots or solutions either real or complex.

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**Roots of Quadratic Equation:**

The values of variables satisfying the given quadratic equation are called its roots. In other words, x = α is a root of the quadratic equation f(x), if f(α) = 0.

The real roots of an equation f(x) = 0 are the x-coordinates of the points where the curve y = f(x) intersect the x-axis.

- One of the roots of the quadratic equation is zero and the other is -b/a if c = 0
- Both the roots are zero if b = c = 0
- The roots are reciprocal to each other if a = c

**Discriminant:**

The term (b^{2} – 4ac) in the quadratic formula is known as the discriminant of a quadratic equation. The discriminant of a quadratic equation reveals the nature of roots.

**Nature of Roots of Quadratic Equation:**

If the value of discriminant = 0 i.e. b^{2} – 4ac = 0 | The quadratic equation will have equal roots i.e. α = β = -b/2a |

If the value of discriminant < 0 i.e. b^{2} – 4ac < 0 | The quadratic equation will have imaginary roots i.e α = (p + iq) and β = (p – iq). Where ‘iq’ is the imaginary part of a complex number |

If the value of discriminant (D) > 0 i.e. b^{2} – 4ac > 0 | The quadratic equation will have real roots |

If the value of discriminant > 0 and D is a perfect square | The quadratic equation will have rational roots |

If the value of discriminant (D) > 0 and D is not a perfect square | The quadratic equation will have irrational roots i.e. α = (p + √q) and β=(p – √q) |

If the value of discriminant > 0, D is a perfect square, a = 1 and b and c are integers | The quadratic equation will have integral roots |

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**Solving Quadratic Equations:**

There are two methods to solve a quadratic equation

- Algebraic Method
- Graphical Method

**Algebraic Method of Solving Quadratic Equations**

__General Form__**:** ax^{2} + bx + c = 0;

x^{2} + bx/a + c/a = 0

⇒ (x + b/2a)^{2} = b^{2}/4a^{2} – c/a

Or, (x + b/2a)^{2} = (b^{2} – 4ac)/4a^{2}

Or, x + b/2a = ± (√b^{2} – 4ac)/2a

⇒ x = [-b ± √(b^{2} – 4ac)]/2a

b^{2} – 4ac = Discriminant (D)

- α = (-b+√D)/2a
- β = (-b – √D)/2a

α+β= -b/a, α.β = c/a

Therefore, the quadratic equation can be written as,

⇒ x^{2} – (α + β)x + (α.β) = 0.

**Quadratic Equation Questions and Answers**

** Question 1)** Find the discriminant of the equation: 3x

^{2}-2x+⅓ = 0.

__Solution__: Here, a = 3, b=-2 and c=⅓

Hence, discriminant, D = b^{2} – 4ac

D = (-2)^{2}-4.3.(⅓)

D = 4-4

D=0

** Question 2)** Solve the quadratic equation 2x

^{2}+ x – 528 = 0, using quadratic formula.

__Solution__: If we compare it with standard equation, ax^{2}+bx+c = 0

a=2, b=1 and c=-528

Hence, by using the quadratic formula:

x=64/4 or x=-66/4

x=16 or x=-33/2

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** Question 3)** Solve the quadratic equation 2x

^{2}+ x – 300 = 0 using factorisation.

__Solution__: 2x^{2}+ x – 300 = 0

2x^{2} – 24x + 25x – 300 = 0

2x (x – 12) + 25 (x – 12) = 0

(x – 12)(2x + 25) = 0

So,

x-12=0; x=12

(2x+25) = 0; x=-25/2 = -12.5

Therefore, 12 and -12.5 are two roots of given equation.

** Question 4)** Check if x(x + 1) + 8 = (x + 2) (x – 2) is in the form of quadratic equation.

__Solution__: Given,

x(x + 1) + 8 = (x + 2) (x – 2)

x^{2}+x+8 = x^{2}-2^{2} [By algebraic identities]

Cancel x^{2} both the sides.

x+8=-4

x+12=0

Since, this expression is not in the form of ax^{2}+bx+c, hence it is not a quadratic equation.

** Question 5) **Find the roots of x

^{2}+ 4x + 5 = 0, if any exist, using quadratic formula.

__Solution__: To check whether there are real roots available for the quadratic equation, we need the find the discriminant value.

D = b^{2}-4ac = 4^{2}-4.1.5 = 16-20 = -4

Since square root of -4 will not give real number. Hence there are no real roots for the given equation.

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** Question 6)** Solve the equation x

^{2}+4x-5=0, using completing the square method.

__Solution__:

x^{2} + 4x – 5 = 0

x^{2}+8/2x-5 = 0

x^{2}+4/2x+4/2x-5 = 0

x^{2}+2x+2x-5 = 0

(x + 2) x + 2 × x – 5= 0

(x + 2) x + 2 × x + 2 × 2 – 2 × 2 – 5= 0

(x + 2) x + (x + 2) × 2 – 2 × 2 – 5 = 0

(x+2) (x+2) -2^{2} – 5 = 0

(x+2)^{2} – 2^{2} – 5 = 0

(x+2)^{2} – 4 – 5 = 0

(x+2)^{2} – 9 = 0

(x+2)^{2} – 3^{2} = 0

(x+2+3) (x+2-3) = 0 [By algebraic identities]

(x+5) (x-1) = 0

Therefore,

x=-5 & x=1

** Question 7)** Find the roots of the equation 2x

^{2}– 5x + 3 = 0 using factorisation.

__Solution__: Given,

2x^{2} – 5x + 3 = 0

2x^{2} – 2x-3x+3 = 0

2x(x-1)-3(x-1) = 0

(2x-3) (x-1) = 0

So,

2x-3 = 0; x = 3/2

(x-1) = 0; x=1

Therefore, 3/2 and 1 are the roots of the given equation.

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** Question 8)** Rahul and Rohan have 45 marbles together. After losing 5 marbles each, the product of the number of marbles they both have now is 124. How to find out how many marbles they had to start with?

__Solution__: Say, the number of marbles Rahul had be x.

Then the number of marbles Rohan had = 45 – x.

The number of marbles left with Rahul after losing 5 marbles = x – 5

The number of marbles left with Rohan after losing 5 marbles = 45 – x – 5 = 40 – x

The product of number of marbles = 124

(x – 5) (40 – x) = 124

40x – x^{2} – 200 + 5x = 124

– x^{2} + 45x – 200 = 124

x^{2} – 45x + 324 = 0

All the details mentioned on this page about Quadratic Equation topic is well written by the team members of recruitmentresult.com Candidates can go through it and practice from the given Quadratic Equation Questions and Answers.

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