Question

Find the value of 'a' in: 9x²+(7a-5)x+25=(3x+5)²

Answer 1

GIVEN :

The equation is $9x^2+(7a-5)x+25=(3x+5)^2$

TO FIND :

The value of a in the given equation.

SOLUTION :

Given equation is $9x^2+(7a-5)x+25=(3x+5)^2$

Now solving the given equation to find the value of a :

$9x^2+(7a-5)x+25=(3x+5)^2$

By using the Algebraic Identity :

$(a+b)^2=a^2+2ab+b^2$

$9x^2+(7a-5)x+25=(3x)^2+2(3x)(5)+(5)^2$ ( here a=3x and b=5 )

By using the property of exponents:

$(ab)^m=a^mb^m$

$9x^2+(7a-5)x+25=3^2x^2+30x+25$

$9x^2+(7a-5)x+25=9x^2+30x+25$

Since the coefficient of $x^2$ and constant term are equal in the above equation we can equate the coefficient of x on the both the sides of the equation we have that

7a-5=30

7a=30+5

7a=35

$a=\frac{35}{7}$

∴ a=5

∴ the value of a in the given equation $9x^2+(7a-5)x+25=(3x+5)^2$ is 5.

∴ the value of a is 5.

Answer 2

Given 9x^+(7a-5)x+25=(3x+5)^ find the value of 'a' is a=5