2x^2+9x+10=0
Ans:
Guide: Product of Binomial⇒ (a+b)^2 = a^2+2ab+b^2
2x^2+9x+10=0 ⇒It is easy to complete the square of an equation if the numerical coefficient of the 1st term is 1. So divide both sides by 2.
x^2+(9/2)x+5=0 ⇒ transpose 5 to the right side so it will become -5.
x^2+(9/2)x=-5 ⇒from the left side of this equation we can use the guide above. So our value of a=1 set aside the variable, when we square the 1 the ans. is always 1. Let's come to (2ab): Our value of 2ab= (9/2) since we already have the value of a as 1 divide (9/2) by 2 to eliminate 2 in 2ab and we can find the value of b.
Our b=(9/4)
x^2+(9/2)x+(9/4)^2 = -5+(9/4)^2 ⇒we add both sides of the equation by b^2.
b=(9/4)
x^2+(9/2)x+(9/4)^2 = -5+(81/16)
[x+(9/4)]^2 = (1/16) ⇒use the guide: the left side. And square both sides of the equation to eliminate the squared of the left side.
x+(9/4)=√(1/16)
x+(9/4)=+ and - of (1/4) ⇒the square root of 1/16 is the positive and negative 1/4
x=+ and - (1/4)-(9/4)
or
x= +(1/4)-(9/4)=-2
x= -(1/4)-(9/4)=-5/2
To check: Use the Quadratic Formula
