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# 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:

**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.**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.**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.**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.**Use Symmetry**: If one root of the quadratic equation is known, exploit the symmetry of the parabola to quickly deduce the other root.**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.**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**^{2 }+ bx + c = 0.

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

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

x^{2}-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 17x ^{2} + 48x – 9 = 0 and 13y^{2} – 32y + 12 = 0**

**Options:**

**A. x < y**

**B. x > y**

**C. x ≤ y**

**D. x ≥ y**

**E. cannot be determined**

**Solution **17x^{2} + 48x – 9 = 0…. (1)

17x^{2} + 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, 13y^{2} – 32y + 12 = 0 ……. (2)

13y^{2} – 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, + x_{2}, + y_{1}, and + y_{2}

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 +x_{1}, +x_{2}, + y_{1}, and + y_{2}

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 9x ^{2} – 36x + 35 = 0 and 2y^{2} – 15y – 17 = 0**

**Options**

**A. x < y B. x > y**

**C. x ≤ y**

**D. x ≥ y**

**E. cannot be determined**

**Solution: **9x^{2} – 36x + 35 = 0……….. (1)

9x^{2} – 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,

2y^{2} – 15y – 17 = 0……. (2)

2y^{2} – 17y + 2y – 17 = 0

(y + 1) (2y – 17) = 0

Therefore, Roots of second equation are 8.5 and -1

Now, compare the roots +x_{1}, + x_{2}, + y_{1}, and – y_{2}

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 x^{2}– 165 = 319……(1)

x^{2}– 165 – 319 = 0

x^{2}= 484 = ±22

y^{2} + 49 = 312…. (2)

y^{2} + 49 -312 = 0

y^{2} = 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 4x^{2} + 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:** 4x^{2} + 18x – 10 = 0……….. (1)

Simplify it, 2x^{2} + 9x – 5 = 0

2x^{2} + 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 -x_{1}, + x_{2}, ± y

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

**Correct option: E**

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- Algebra – Questions | Formulas | How to Solve Quickly | Tricks & Shortcuts
- Linear Equations – Questions | Formulas | How to Solve Quickly | Tricks & Shortcuts

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Tricks & Shortcuts - Linear Equations –

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