A quadratic equation takes the form ax2 + bx + c = 0, representing a second-degree polynomial equation, where 'a', 'b', and 'c' are constants, and 'x' is the variable. It is essential for 'a' to be nonzero to validate it as a genuine quadratic equation.

## Methods for finding roots of Quadratic Equation

The roots or solutions of a quadratic equation are the values of ‘x’ that satisfy the equation, making it true. A quadratic equation can have two real roots, two complex roots, or one real root (in case of a perfect square). The number of roots is determined by the value of the discriminant (Δ) given by:

Δ = b2 −4ac

• If Δ > 0, the quadratic equation has two distinct real roots.
• If Δ = 0, the quadratic equation has one real root (repeated or equal roots).
• If Δ < 0, the quadratic equation has two complex roots (conjugate pairs).

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

• Factoring: If the equation can be factored, the roots can be directly determined from the factors.
• Quadratic Formula: The quadratic formula is a general method to find the roots of any quadratic equation and is given by:
where the ± sign accounts for the two possible solutions.
• Completing the Square: This method involves transforming the quadratic equation into a perfect square form, from which the roots can be easily obtained.

• If p+ √q is a root of a quadratic equation, then its other root is p-√q.
• When D ≥ 0, then rx2+ dx + z can be expressed as a product of two linear factors.
• If α and β are the roots of rx2+ dx + z, then we can write it as: x2 – (α + β)x + α β = 0.

Examples

Example1: If x2 - 3x + 1 = 0, find the value of x + 1/x,
(a) 0
(b) 3
(c) 2
(d) 1
Ans:
(b)
Given equation is
x2 - 3x + 1 = 0 ⇒ x2 + 1 = 3x
⇒ x2 + 1/x = 3
⇒ x2/x + 1/x = 3
∴ x + 1/x = 3

Example 2: For what value of k, the equation x2 + 2(k - 4) x + 2k = 0 has equal roots ?
(a) 6 and 4
(b) 8 and 2
(c) 10 and 4
(d) 12 and 2
Ans:
(b)
Given equation is
x2 + 2(k - 4)x + 2k = 0
On comparing with ax2 + bx + c = 0
Here, a = 1, b = 2(k -4), c = 2k
Since, the root are equal, we have D = 0.
b2 - 4ac = 0
∴ 4(k - 4)2 - 8k = 0
4(k2 + 16 - 8k) - 8k = 0
⇒ 4k2 + 64 - 32k - 8k = 0
⇒ 4k2 - 40k + 64 = 0
⇒ k2 - 10k + 16 = 0
⇒ k2 - 8k - 2k + 16 = 0
⇒ k(k - 8) -2 (k - 8) = 0
⇒ (k - 8) (k - 2) = 0
Hence, the value of k 8 or 2.

Example 3: If α and β are the roots of the equation 4x2 - 19x + 12 = 0, find the equation having the roots 1/α and 1/β
(a) 4x+ 19 + 12 = 0
(b) 12x2 - 19x + 4 = 0
(c) 12x2 + 19x + 4 = 0
(d) 4x2 + 19x - 12 = 0
Ans:
(b)
Given equation is 4x2 - 19x + 12 = 0
Let given equation having the roots 1/α and 1/β,
Then required equuation is
12x2 - 19x + 4 = 0

Example 4: If one of the roots of quadratic equation 7y2 - 50y + k = 0 is 7, then what is the value of k ?
(a) 7
(b) 1
(c) 50/7
(d) 7/50
Ans:
(a)
7y2 - 50y + k = 0
If one root is 7, then it will satisfy the equation i.e putting y = 7 in equation
7 x (7)2 - 50 x 7 + k = 0
⇒ 7 x 49 - 350 + k = 0
⇒ 343 - 350 + k = 0
∴ k = 7

Example 5: The quadrictic equation whose roots are 3 and -1, is
(a) x2 - 4x + 3 = 0
(b) x2 - 2x - 3 = 0
(c) x2 + 2x - 3 = 0
(d) x2 + 4x + 3 = 0
Ans:
(b)
Given that, the roots of the quadrictic equation are 3 and -1.
Let α = 3 and β = -1
Sum of roots = α + β = 3 - 1 = 2
Products of roots = α . β = (3) (-1) = -3
x2 - (α + β)x + α β = 0
⇒ x2 - (2)x + (-3) = 0
⇒ x2 - 2x - 3 = 0

The document Solved Examples: Quadratic Equations | Quantitative Aptitude for SSC CGL is a part of the SSC CGL Course Quantitative Aptitude for SSC CGL.
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## Quantitative Aptitude for SSC CGL

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## FAQs on Solved Examples: Quadratic Equations - Quantitative Aptitude for SSC CGL

 1. What is a quadratic equation?
Ans. A quadratic equation is a second-degree polynomial equation in a single variable, typically written as ax^2 + bx + c = 0, where a, b, and c are constants and a ≠ 0.
 2. How do you solve a quadratic equation by factoring?
Ans. To solve a quadratic equation by factoring, set the equation equal to zero and try to factor the quadratic expression. Once factored, set each factor equal to zero and solve for the variable to find the solutions.
 3. Can a quadratic equation have more than two solutions?
Ans. No, a quadratic equation can have at most two solutions. This is because a quadratic equation is a second-degree polynomial and, by the fundamental theorem of algebra, a polynomial of degree n can have at most n solutions.
 4. What is the quadratic formula and how is it used to solve quadratic equations?
Ans. The quadratic formula is given by x = (-b ± √(b^2 - 4ac)) / (2a), where a, b, and c are the coefficients of the quadratic equation ax^2 + bx + c = 0. The quadratic formula can be used to directly find the solutions of a quadratic equation without factoring.
 5. Are there any real-life applications of quadratic equations?
Ans. Yes, quadratic equations have numerous real-life applications. They are used in physics to describe projectile motion, in engineering to model parabolic trajectories, in finance for profit and loss calculations, and in biology to study population growth, among many other applications.

## Quantitative Aptitude for SSC CGL

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