# 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. ### 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

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### 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

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