How To Solve Quadratic Equations Quickly

How to Solve Quadratic Equation Questions Quickly

When it comes to swiftly solving quadratic equations, a few key tips can significantly expedite the process. Start by recognizing opportunities for efficient factoring, which can unveil the equation’s roots effortlessly. Should factoring pose a challenge, leverage the quadratic formula—a potent tool providing precise solutions.

how to solve quadriatic equation questions quickly

Methods to solve Quadratic Equation Quickly

Solving quadratic equations quickly involves employing various methods that streamline the process and provide efficient solutions. Here are some effective methods:

  1. Factoring: If possible, factor the quadratic equation into two binomials and set each binomial equal to zero. This method is particularly efficient when the equation is easily factorable.

  2. Quadratic Formula: Utilize the quadratic formula to directly find the solutions of the equation. The formula is x = (-b ± √(b² – 4ac)) / 2a, where “a,” “b,” and “c” are the coefficients of the quadratic equation. This method is reliable and applicable to all quadratic equations.

  3. Completing the Square: Transform the quadratic equation into a perfect square trinomial by adding and subtracting a suitable constant. This method is useful for equations that are not easily factorable, and it helps simplify the process of finding the roots.

  4. Graphical Analysis: Graph the quadratic equation on a coordinate plane and determine the points where the graph intersects the x-axis. This method provides an approximate visual representation of the roots.

  5. Use Symmetry: If one root of the quadratic equation is known, exploit the symmetry of the parabola to quickly deduce the other root.

  6. Shortcut Techniques: For specific types of quadratic equations, such as those with special patterns (perfect squares, difference of squares), you can use shortcut techniques to simplify the solving process.

  7. Mental Math: Develop mental math skills to perform calculations quickly and accurately. Simplify coefficients and terms mentally to expedite the calculations.

How To Solve Quadratic Equation & Definition

  • A quadratic equation is an equation having the form ax+ bx + c = 0.

Where,  x is the unknown variable and a, b, c are the numerical coefficients.

Example : Solve for x : x2-3x-10 = 0
Solution :  Let us express -3x as a sum of -5x and +2x.

 x2-5x+2x-10 = 0

x(x-5)+2(x-5) = 0

(x-5)(x+2) = 0

 x-5 = 0 or x+2 = 0

 x = 5 or x = -2

Type 1: Solving Quadratic Equations Questions Quickly

  • In each of these questions, two equations are given. You have to solve these equations and find out the values and relation of between x and y.

Question 1  Solve for the equations 17x2 + 48x – 9 = 0 and 13y2 – 32y + 12 = 0

Options:

A. x < y

B. x > y

C. x ≤ y

D. x ≥ y

E. cannot be determined

Solution     17x2 + 48x – 9 = 0…. (1)

17x2 + 51x – 3x – 9 = 0

(x+3) (17x – 3) = 0

Therefore, Roots of first equation are -3 and \frac{3}{17}

We know that, if sign given in the equation is + and – then their sign of roots is – and + respectively.

Therefore, the roots of the equation are – 3 and – \frac{3}{17}

Now, 13y2 – 32y + 12 = 0 ……. (2)

13y2 – 26y – 6y + 12 = 0

(y-2) (13y – 6) = 0

Therefore, Roots of second equation are 2 and \frac{6}{3}

We know that, If sign given in the equation is – and – then their sign of roots is + and + respectively.

Therefore, the roots of the equation are +2 and + \frac{6}{13}

Now, compare the roots – x1, + x2, + y1, and + y2

It means y>x

Correct option: A

Question 2. Solve for the equations \mathbf{\sqrt{500}x =\sqrt{420}}  and \mathbf{\sqrt{260}y – \sqrt{200} = 0  }

Options

A. x < y

B. x > y

C. x ≤ y

D. x ≥ y

E. cannot be determined

Solution:    \sqrt{500}x -\sqrt{420} = 0 …… (1)

\sqrt{500}x =\sqrt{420} 

500 x = 420

x =\frac{420}{500}

x = \frac{210}{250}

x = 0.84

Now, \sqrt{260}y -\sqrt{200} = 0

\sqrt{260}y = \sqrt{200} 

260y = 200

y = \frac{200}{260}

y = \frac{100}{130}

y = \frac{5}{9}

y = 0.76

On comparing x and y, it is clear that x > y

Correct option: B

Question 3. Solve for the equations    x ^{2} – 11x + 24 = 0  and     2y^{2}– 9y + 9 = 0

Options

A. x < y

B. x > y

C. x ≤ y

D. x ≥ y

E. cannot be determined

Solution: x^{2} – 11x + 24 = 0 ……….. (1)

                 x^{2} – 8x-3x + 24 = 0

                (x – 8) (x – 3) = 0

Therefore, Roots of first equation are 8 and 3

We know that, If sign given in the equation is  – and – then their sign of roots is + and +

Therefore, the roots of the equation are +8 and + 3

Now,  2y^{2}– 9y + 9 = 0 ……. (2)

2y^{2}– 6y – 3y+ 9 = 0

(y-3) (2y – 3) = 0

Therefore, Roots of second equation are 3 and 1.5

We know that, If sign given in the equation is – and + then their sign of roots is + and +

Therefore, the roots of the equation are + 3 and + 1.5

Now, compare the roots +x1, +x2, + y1, and + y2

It means x ≥ y

Correct option: D

Prime Course Trailer

Related Banners

Get PrepInsta Prime & get Access to all 200+ courses offered by PrepInsta in One Subscription

Type 2: How To Solve Quadratic Equation Questions

  • When relation cannot be determined

Question 4. Solve for equations 9x2 – 36x + 35 = 0 and 2y2 – 15y – 17 = 0

Options

A. x < y B. x > y

C. x ≤ y

D. x ≥ y

E. cannot be determined

Solution:    9x2 – 36x + 35 = 0……….. (1)

9x2 – 21x – 15x + 35 = 0

(3x – 7) (3x – 5) = 0

Therefore, Roots of first equation are \frac{7}{3} and \frac{5}{3}

We know that, If sign given in the equation is  – and – then their sign of roots is + and + respectively

Therefore, the roots of the equation are +1.66 and +2.33

Now,

2y2 – 15y – 17 = 0……. (2)

2y2 – 17y + 2y – 17 = 0

(y + 1) (2y – 17) = 0

Therefore, Roots of second equation are 8.5 and -1

Now, compare the roots +x1, + x2, + y1, and – y2

It means , we cannot find any relation between x and y

Correct option: E

Question 5. Solve for equations  x^{2} –  165 = 319 and y^{2} + 49 = 312  

Options

A. x < y B. x > y

C. x ≤ y

D. x ≥ y

E. cannot be determined

Solution      x2– 165 = 319……(1)

x2– 165 – 319 = 0

x2= 484  = ±22

y2 + 49 = 312…. (2)

y2 + 49 -312 = 0

y2 = 263 y = ± 16.21

Now compare, x and y It means, we cannot find any relation between x and y

Correct option: E

Question 3. Solve for Equations   4x2 + 18x – 10 = 0 and  y^\frac{2}{5}(\frac{25}{y})^\frac{8}{5} = 0

Options:

A. x < y B. x > y

C. x ≤ y

D. x ≥ y

E. cannot be determined

Solution:     4x2 + 18x – 10 = 0……….. (1)

Simplify it, 2x2 + 9x – 5 = 0

2x2 + 10x – x – 5 = 0

(x + 5) (2x – 1) = 0

Therefore, Roots of first equation are 5 and 0.5

We know that, if sign given in the equation is + and – then their sign of roots is – and +

Therefore, the roots of the equation are – 5 and +0.5

Now, y^\frac{2}{5} – (\frac{25}{y})^\frac{8}{5} = 0……. (2)

y^\frac{2}{5}  –  (\frac{5^2}{y})^\frac{8}{5} = 0

y = ± 5

Now, compare the roots -x1, + x2, ± y

It means, we cannot find any relation between x and y

Correct option: E

Also Check Out

Get over 200+ course One Subscription

Courses like AI/ML, Cloud Computing, Ethical Hacking, C, C++, Java, Python, DSA (All Languages), Competitive Coding (All Languages), TCS, Infosys, Wipro, Amazon, DBMS, SQL and others

Checkout list of all the video courses in PrepInsta Prime Subscription

Checkout list of all the video courses in PrepInsta Prime Subscription