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# Tips And Tricks And Shortcuts For Quadratic Equations

## Unlocking Tips, Tricks, and Shortcuts for Quick Solutions

Explore an invaluable resource for effortlessly tackling quadratic equations with finesse. Our comprehensive guide unveils a treasure trove of proven strategies, smart shortcuts, and expert tips that will elevate your quadratic equation-solving prowess.

**Tips and shortcuts to solve the Quadratic Equation questions**

- In order to find the sign of roots we use sign given in the equation.

Sign of coefficient ‘x’ | Sign of coefficient ‘y’ | Sign of roots |
---|---|---|

+ | + | – – |

+ | – | – + |

– | + | + + |

– | – | + – |

For example: x^{2} + 3x – 4 = 0

The factors are 4 and 1

Now will find the sign of roots. According to the table, if sign given in the equation is + and – then their sign of roots is – and +. Therefore, the roots of the equation are -4 and 1

### Type 1: Tricks and Shortcuts for Quadratic Questions

- When one equation has positive roots and other has negative, or equations has same sign roots, then compare the roots and find the relation between them.

**Question 1:**

** Find the correct option given below by solving the following equation ****x ^{2} – 7x + 10 = 0 and **

**y**

^{2}+ 8y + 15 = 0**Options:**

**A. x < y**
**B. x > y**
**C. x ≤ y**
**D. x ≥ y**
**E. cannot be determined**

**Solution:** x^{2} – 7x + 10 = 0…. (1)

Roots of first equation are 5 and 2

Using the table above, we will find the sign of roots. If sign given in the equation is – and + then their sign of roots is + and +

Therefore, the roots of the equation are 5 and 2

Now, y^{2} + 8y + 15 = 0 ……. (2)

Roots of second equation are 3 and 5

If sign given in the equation is + and + then their sign of roots is – and –

Therefore, the roots of the equation are -3 and – 2

Now, here roots are +x1, + x2, – y1, and – y2

This clearly shows that X roots are positive and Y roots are negative. So, x> y

**Correct option: B**

### Type 2: Tips and Tricks To Solve Quadratic Equations Questions

- When roots of the equation are positive and negative and we cannot find the relation between them.

**Question 2:**

** Find the correct option given below by solving the following equation **6x^{2} + 11x – 35 = 0 and 6y^{2} + 5y – 6 = 0

**Options:**

**A. x < y**
**B. x > y**
**C. x ≤ y**
**D. x ≥ y**
**E. cannot be determined**

**Solution: **6x^{2} + 11x – 35 = 0…. (1)

Roots of first equation are 3.5 and 1.66

Using the table above, we will find the sign of roots. If sign given in the equation is + and – then their sign of roots is – and +

Therefore, the roots of the equation are – 3.5 and 1.66

Now, 6y^{2} + 5y – 6 = 0 ……. (2)

Roots of second equation are 1.5 and 0.66

If sign given in the equation is + and – then their sign of roots is – and +

Therefore, the roots of the equation are – 1.5 and 0.66

Now, here roots are -x_{1}, + x_{2}, – y_{1}, and + y_{2}

It means x_{2}> – y_{1} and –x_{1}< + y_{2}

So, we cannot determine which roots are greater.

**Correct option: E**

### Type 3: Tips and Tricks and Shortcuts for Quadratic Questions

- When we can find a equation where x is a variable and a, b, and c represent constants and D, i.e., Discriminant.

The discriminant is the part of the quadratic formula underneath the square root symbol: b²-4ac.
**
Question 3: **

**Find the Discriminant and roots of the following equation: 3x ^{2 }− 5x − 7 = 0.**

**Solution:**

Let us find Discriminant, D = \sqrt{B^{2}-4AC}

D = \sqrt{(-5)^{2}-4(3)(-7)}

D = 109

Therefore, x = \frac{-b \pm \sqrt{}D}{2a}

x = \frac{5 \pm \sqrt{}109}{6}

x = \frac{5 \pm 10.44}{6}

x = \frac{15.44}{6} => x = 2.57 , and

x = \frac{-5.44}{6} => x = – 0.90

So the two roots are 2.57 and -0.90

**Some more Questions:**

**Question 4:**

**Which method is used to solve quadratic equations that cannot be easily factored or involve complex coefficients? **

**A) Factoring **

**B) Quadratic Formula **

**C) Completing the Square **

**D) Estimation**

**Explanation:**

The correct answer is option B. The quadratic formula is a universal method used to solve any quadratic equation, regardless of its coefficients or factorability. It provides precise solutions by directly calculating the roots.

**Question 5:**

**For the quadratic equation 3x² + 6x – 9 = 0, what are the roots? **

**A) x = 1 **

**B) x = -1 **

**C) x = -3 **

**D) x = 3**

**Explanation:**

The correct answer is option C. Divide the entire equation by 3 to simplify it: x² + 2x – 3 = 0. This equation can be factored as (x – 1)(x + 3) = 0. Setting each factor to zero gives x – 1 = 0 and x + 3 = 0. Solving for “x” in both equations yields x = 1 and x = -3. Therefore, the roots are x = -3 and x = 1.

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

- Algebra –

Questions |

Formulas |

How to Solve Quickly |

Tricks & Shortcuts - Linear Equations –

Questions |

Formulas |

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Tricks & Shortcuts

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