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Solve ax²-9(a+b)x+(2a²+5ab+2b²)=0
By BYJU'S Exam Prep
Updated on: September 25th, 2023
Value of x = (a + 2b)/3 or x = (2a + b)/3 for Equation ax²-9(a+b)x+(2a²+5ab+2b²)=0.
Consider, 9x²-9(a+b)x+(2a²+5ab+2b²)=0
First, we have to consider (2a²+5ab+2b²) and factorize it:
= 2a²+4ab+ab+2b²
The above equation can be written as:
= 2a(a+2b)+b(a+2b)
Now we get:
= (2a+b)(a+2b)
Multiplying the above equation we get:
9x² – 9(a + b)x + (2a² + 5ab + 2b²) = 0
Taking common in the above equation we get:
9x² – 9(a + b)x + [(2a + b)(a + 2b)] = 0
9x² – 3*3(a + b)x + [(2a + b)(a + 2b)] = 0
9x² – 3(3a + 3b)x + [(2a + b)(a + 2b)] = 0
Substituting the values in the above equation we get:
9x² – 3[(a + 2b) + (2a + b)]x + [(a + 2b)(2a + b)] = 0
9x² – 3(a + 2b)x – 3(2a + b)x + (a + 2b)(2a + b) = 0
Taking 3x common in the above equation we get:
3x[3x – (a + 2b)] – (2a + b)[3x – (a +2b)] = 0
The above equation becomes:
[3x – (a + 2b)][3x – (2a + b)] = 0
[3x – (a + 2b)] = 0 or [3x – (2a + b)] = 0
Therefore:
3x = (a + 2b) or 3x = (2a + b)
Table of content
Factorisation
An algebraic expression is factorised when it is written as the product of its factors. These variables, factors, or algebraic expressions could be present.
The algebraic expressions can be factored in using one of four techniques.
- Common factors method
- Factorisation using identities
- Regrouping terms method
- Factors of the form (x+a) (x+b)
Summary:
Solve ax²-9(a+b)x+(2a²+5ab+2b²)=0
On solving equation ax²-9(a+b)x+(2a²+5ab+2b²)=0, we get the value of x is x = (a + 2b)/3 or x = (2a + b)/3. It is solved using the common factors method of algebraic expressions.
