## Tips and shortcuts to solve the Quadratic Equation questions

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

For example: x2 + 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 while the other has negative roots, or when both equations have roots of the same sign, compare the roots to determine the relationship between them.

Q1: Find the correct option given below by solving the following equation  x2 – 7x + 10 = 0 and y2 + 8y + 15 = 0
(a) x < y
(b) x > y
(c) x ≤ y
(d) x ≥ y
Ans:
(b)
x2 – 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, y2 + 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

### 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.

Q2: Find the correct option given below by solving the following equation 6x2 + 11x – 35 = 0 and  6y2 + 5y – 6 = 0
(a) x < y
(b) x > y
(c) x ≤ y
(d) cannot be determined
Ans:
(d)
6x2 + 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, 6y2 + 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 -x1, + x2, – y1, and + y2
It means x2> – y1 and –x1< + y2
So, we cannot determine which roots are greater.

### 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.

Q3: Find the Discriminant and roots of the following equation: 3x2 − 5x − 7 = 0.
Sol:

Let us find Discriminant,

D = 109

So the two roots are 2.57 and -0.90

Q4: Which method is used to solve quadratic equations that cannot be easily factored or involve complex coefficients?
(a) Factoring
(c) Completing the Square
(d) Estimation
Ans:
(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.

Q5: For the quadratic equation 3x² + 6x – 9 = 0, what are the roots?
(a) x = 1
(b) x = -1
(c) x = -3
(d) x = 3
Ans:
(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.

The document Tips and Tricks: 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 Tips and Tricks: Quadratic Equations - Quantitative Aptitude for SSC CGL

 1. What is a quadratic equation?
Ans. A quadratic equation is a polynomial equation of the second degree, which means it has a variable raised to the power of two. It can be represented in the form ax^2 + bx + c = 0, where a, b, and c are coefficients, and x is the variable.
 2. How do you solve a quadratic equation using the quadratic formula?
Ans. To solve a quadratic equation using the quadratic formula, first identify the values of a, b, and c from the given equation. Then, substitute these values into the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a. Finally, simplify the equation to find the values of x.
 3. Are there any shortcuts to solve quadratic equations?
Ans. Yes, there are a few shortcuts to solve quadratic equations. One such shortcut is factoring, where you try to factorize the equation and solve for x. Another shortcut is completing the square, where you manipulate the equation to create a perfect square trinomial, making it easier to solve. Additionally, using the quadratic formula is considered a shortcut as it provides a direct way to find the solutions.
 4. Can all quadratic equations be solved using shortcuts?
Ans. While most quadratic equations can be solved using shortcuts like factoring, completing the square, or the quadratic formula, there are some cases where the solutions may involve complex numbers or be impossible to find. However, these cases are relatively rare and most quadratic equations encountered in exams or daily life can be solved using the available shortcuts.
 5. What are some tips for solving quadratic equations efficiently?
Ans. Here are some tips for solving quadratic equations efficiently: - Always check if the equation can be factored before using other methods. - Use the quadratic formula as a last resort if factoring or completing the square is not possible. - Simplify the equation before applying any method to make the calculations easier. - Pay attention to the signs and coefficients while solving to avoid mistakes. - Practice solving different types of quadratic equations to improve your speed and accuracy.

## Quantitative Aptitude for SSC CGL

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