SOLVING QUADRATIC EQUATIONS BY FACTORING
About "Solving quadratic equations by factoring"
Solving quadratic equations by factoring :
Here we are going to see how to solve quadratic equation in a simple way.
Generally we have two types of quadratic equation
Solving quadratic equations by factoring with a leading coefficient of 1
In a quadratic "Leading coefficient" means "coefficient of x²".
(i) If the leading coefficient is 1, we have to take the constant term and we have to split it into two parts.
(ii) The product of two parts must be equal to the constant term and the simplified value must be equal to the middle term (or) x term.
(iii) Now we have to write these numbers in the form of (x + a) and (x +b)
(iv) By equating the factors equal to zero. We may find the value of x. solving quadratic equations by factoring
Solving quadratic equations by factoring - Examples
Example 1 :
Solve x² + 17 x + 60 = 0
Solution :
In the first step we are going to check whether the coefficient of x² is 1 or not.
Since it is 1, we need to take the last number. That is 60 and find the factors of 60.
Since the given equation is in the form ax2 + bx + c, we have to put positive sign for both factors.
Here,
10 x 6 = 60 but 10 + 6 = 16 not 17
15 x 4 = 60 but 15 + 4 = 19 not 17
12 x 5 = 60 and 12 + 5 = 17
2 x 30 = 60 but 2 + 30 = 32 not 17
(x + 12) (x + 5) are the factors
x + 12 = 0 x + 5 = 0
x = -12 x = -5
Solution is {-12,-5}
This is just example of solving quadratic equations by factoring. If you need more example problems of solving quadratic equations by factoring please click the below link.
Solving quadratic equations by factoring when leading coefficient isn't 1
(i) If it is not 1, then we have to multiply the coefficient of x² by the constant term and we have to split it as two parts.
(ii) The product of two parts must be equal to the constant term and the simplified value must be equal to the middle term (or) x term.
(iii) Divide the factors by the coefficient of x². Simplify the factors by the coefficient of x² as much as possible.
(iv) Write the remaining number along with x.
Example problem of solving quadratic equations by factoring
Example 2 :
Solve 2 x² + x - 6 = 0
Solution :
To factor this quadratic equation we have to multiply the coefficient of x² by the constant term
So that we get -12, now we have to split -12 as the multiple of two numbers.
Since the last term is having negative sign, we have to put negative sign for the least number.
Now we have to divide the two numbers 4 and -3 by the coefficient of x² that is 2. If it is possible we may simplify, otherwise we have to write the numbers along with x.
(x + 2) (2x - 3) = 0
x + 2 = 0 2 x - 3 = 0
x = -2 2 x = 3
x = 3/2
This is just one example problem to show solving quadratic equations by factoring. If you want more example please click the below link.
More example problems
Problem 1 :
Solve by factoring method (2x + 3)² - 81 = 0
Solution :
(2x + 3)² - 81 = 0
Now we are going to compare (2x + 3)² with the algebraic identity (a+b)² = a² + 2 a b + b²
(2x + 3)² - 81 = 0
(2 x)² + 2 (2 x) (3) + 3² - 81 = 0
4 x² + 12 x + 9 – 81 = 0
4 x² + 12 x – 72 = 0
Now we are going to divide the whole equation by 4
x² + 3 x – 18 = 0
(x + 6) (x – 3) = 0
x + 6 = 0 x – 3 = 0
x = -6 x = 3
Alternate way:
(2x + 3)² - 81 = 0
(2x + 3)² = 81
(2 x + 3) = √81
(2 x + 3) = ± 9
2 x + 3 = 9 2 x + 3 = - 9
2 x = 9 – 3 2 x = - 9 – 3
2 x = 6 2 x = -12
x = 6/2 x = -12/2
x = 3 x = -6
Verification:
(2x + 3)² - 81 = 0
if x = 3
(2 (3) + 3)² - 81 = 0
(6 + 3)² - 81 = 0
9² - 81 = 0
81 - 81 = 0
0 = 0
if x = -6
(2 (-6) + 3)² - 81 = 0
(-12 + 3)² - 81 = 0
(-9)² - 81 = 0
81 - 81 = 0
0 = 0
Now let us see the next example problem of the concept "solving quadratic equations by factoring".
Problem 2 :
Solve by factoring method 3 x² – 5 x – 12 = 0
Solution :
3 x² – 5 x – 12 = 0
3 x² – 9 x + 4 x – 12 = 0
3 x (x – 3) + 4 (x – 3) = 0
(3 x + 4) (x - 3) = 0
3 x + 4 = 0 x – 3 = 0
3 x = -4 x = 3
x = -4/3
Verification:
3 x² – 5 x – 12 = 0
if x = 3
3 (3)² – 5 (3) – 12 = 0
3(9) - 15 - 12 = 0
27 - 27 = 0
0 = 0
if x = -4/3
3 (-4/3)² – 5 (-4/3) – 12 = 0
(16/3) + (20/3) - (36/3) = 0
(16 + 20 - 36)/3 = 0
(36 - 36)//3 = 0
0/3 = 0
0 = 0
Now let us see the next example problem of the concept "solving quadratic equations by factoring".
Problem 3:
Solve by factoring method √5 x² + 2 x – 3√5 = 0
Solution :
√5 x² + 5 x - 3 x – 3√5 = 0
√5 x (x + √5) – 3 (x + √5) = 0
(√5 x – 3) (x + √5) = 0
√5 x – 3 = 0 x + √5 = 0
√5 x = 3 x = - √5
x = 3/√5
Verification:
√5 x² + 2 x – 3√5 = 0
if x = 3/√5
√5 (3/√5)² + 2 (3/√5) – 3√5 = 0
√5(9/5) + (6/√5) – 3√5 = 0
(9 √5/5) + (6√5/5) - (15√5)/5 = 0
(9 √5+ 6√5 - 15√5)/5 = 0
(15 √5 - 15 √5)/5 = 0
0/5 = 0
0 = 0
Now let us see the next example problem of the concept "solving quadratic equations by factoring".
Problem 4 :
Solve by factoring method 3 (x² – 6) = x (x + 7) – 3
Solution :
3 x² – 18 = x² + 7 x – 3
3x² - x² – 7 x – 18 + 3 = 0
2 x² – 7 x – 15 = 0
2 x² – 10 x + 3 x – 15 = 0
2 x (x – 5) + 3( x – 5) = 0
(2 x + 3) (x – 5) = 0
2 x + 3 = 0 x – 5 = 0
2x = -3 x = 5
x = -3/2
Verification:
2 x² – 7 x – 15 = 0
if x = -3/2
2 (-3/2)² – 7(-3/2) – 15 = 0
2 (9/4) + 21/2 – 15 = 0
(9/2) + (21/2) - 15 = 0
(9 + 21 - 30)/2 = 0
(30 - 30)/2 = 0
0 = 0
if x = 5
2 x² – 7 x – 15 = 0
2 (5)² - 7(5) - 15 = 0
2 (25) - 35 - 15 = 0
50 - 50 = 0
0 = 0
Now let us see the next example problem of the concept "solving quadratic equations by factoring".
Problem 5 :
Solve by factoring method 3 x – (8/x) = 2
Solution :
(3 x² – 8)/x = 2
(3 x² – 8) =2 x
3 x² - 2 x – 8 = 0
3 x² - 6 x + 4 x – 8 = 0
3 x (x – 2) + 4 (x – 2) = 0
(3 x + 4) (x – 2) = 0
3 x + 4 = 0 x – 2 = 0
3x = -4 x = 2
x = -4/3
Verification:
3 x² - 2 x – 8 = 0
if x = -4/3
3 (-4/3)² - 2(-4/3) – 8 = 0
(16/3) + (8/3) - 8 = 0
(16 + 8 - 24)/3 = 0
(24 - 24)/3 = 0
0 = 0
if x = 2
3 (2)² - 2 (2) – 8 = 0
3 (4) - 4 - 8 = 0
12 - 4 - 8 = 0
12 - 12 = 0
0 = 0
Now let us see the next example problem of the concept "solving quadratic equations by factoring".
Problem 6 :
x + (1/x) = (26/5)
Solution :
(x² + 1)/x = 26/5
5 (x² + 1) = 26 x
5 x² + 5 = 26 x
5 x² - 26 x + 5 = 0
5 x² - 25 x – x + 5 = 0
5 x (x -5) – 1(x – 5) = 0
(5 x – 1) (x – 5) = 0
5 x – 1 = 0 x – 5 =0
5 x = 1 x = 5
x = 1/5
Verification:
5 x² - 26 x + 5 = 0
if x = 1/5
5(1/5)² - 26(1/5) + 5 = 0
(1/5) - (26/5) + 5 = 0
(1 - 26 + 25)/5 = 0
(26 - 26)/5 = 0
0/5 = 0
0 = 0
if x = 5
5 (5)² - 26(5) + 5 = 0
5 (25) - 130 + 5 = 0
125 + 5 - 130 = 0
130 - 130 = 0
0 = 0
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