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There are many ways to solve the quadratic equations.We would be studying detail about each method with some solve examples

In this method of Solving quadratic equations by factoring, we factorize the Quadratic equation by splitting the middle term b

In $ax^2+bx+c=0$

Step 1: Arrange the equation so that coefficient a is positive. If a is negative,multiply each term by -1 to make it positive
Step 2: Find out the product of the coefficient a and c i.e ac.
Step 3: if ac is positive then Find the factor of the product ac such that it adds to form the coefficient b. If b is negative,you can place factors with negative sign and if b is positive, place the factors with positive sign
Step 4: if ac is negative,then Find the factor of the product ac such that it subtract to form the coefficient b. If b is positive, then the larger factor is positive. If b is negative, then the larger factor is negative.
Table below explain the step 3 and 4
 b >0 b <0 ac > 0 Both the factors > 0 Both the factors < 0 ac < 0 Both the factor will have opposite sign i.e. one positive and one negative. Positive factor will have larger value. Both the factor will have opposite sign i.e. one positive and one negative. negative factor will have larger value.
Step 5:We can write the equation in split form and factorize the equation.
Step 6: Roots of the equation can be find equating the factors to zero

Solving quadratic equations by factoring Examples

Example -1
$6x^2-x-2=0$
Solution
Step 1 First we need to multiple the coefficient a and c.In this case =6X-2=-12
Step 2 Now the product is negative,so Splitting the middle term so that multiplication is -12 and difference is the coefficient b. Here in this case factor 3, 4 are present,so we can take 3 and -4
$6x^2 +3x-4x-2=0$
$3x( 2x+1) -2(2x+1)=0$
$(3x-2) (2x+1)=0$
step 3 Roots of the equation can be find equating the factors to zero
$3x-2=0$ => $x=\frac {3}{2}$
$2x+1=0$ => $x=\frac {-1}{2}$

Example -2
$x^2 + 7x + 6=0$
Solution
Step 1 First we need to multiple the coefficient a and c.In this case =1X6=6
Step 2 Now the product is positive,so Splitting the middle term so that multiplication is 6 and addition is the coefficient b. Here in this case factor 1, 6 are present,so we can take 1 and 6
$x^2 +6x+x-2=0$
$x( x+1) +6(x+1)=0$
$(x+1) (x+6)=0$
step 3 Roots of the equation can be find equating the factors to zero
$x+1=0$ => $x=-1$
$x+6=0$ => $x=-6$

Practice Questions
• $x^{2} - 13 x + 40 = 0$
• $x^{2} + 2 x - 48 = 0$
• $x^{2} - 8 x + 7 = 0$
• $x^{2} - 1 = 0$
• $x^{2} + 3 x - 54 = 0$
• $x^{2} - 15 x + 56 = 0$
• $x^{2} + 2 x - 3 = 0$
• $x^{2} - 7 x + 6 = 0$
• $x^{2} - 2 x - 63 = 0$
• $x^{2} + x - 42 = 0$

Reference Books for class 10

Given below are the links of some of the reference books for class 10 math.

You can use above books for extra knowledge and practicing different questions.

Practice Question

Question 1 What is $1 - \sqrt {3}$ ?
A) Non terminating repeating
B) Non terminating non repeating
C) Terminating
D) None of the above
Question 2 The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is?
A) 19.4 cm3
B) 12 cm3
C) 78.6 cm3
D) 58.2 cm3
Question 3 The sum of the first three terms of an AP is 33. If the product of the first and the third term exceeds the second term by 29, the AP is ?
A) 2 ,21,11
B) 1,10,19
C) -1 ,8,17
D) 2 ,11,20

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