How to obtain the constant to be added to both sides of the quadratic equation so that the left side will form a perfect square trinomial? Please answer correctly.

Step-by-step explanation:

According to steps 2 and 4. The second-order polynomial must be added by -c−c and b^{2}b

2

to create a perfect square trinomial.

Step-by-step explanation:

Let consider a second-order polynomial of the form a\cdot x^{2} + b\cdot x + c = 0a⋅x

2

+b⋅x+c=0 , \forall \,x \in\mathbb{R}∀x∈R . The procedure is presented below:

1) a\cdot x^{2} + b\cdot x + c = 0a⋅x

2

+b⋅x+c=0 (Given)

2) a\cdot x^{2} + b \cdot x = -ca⋅x

2

+b⋅x=−c (Compatibility with addition/Existence of additive inverse/Modulative property)

3) 4\cdot a^{2}\cdot x^{2} + 4\cdot a \cdot b \cdot x = -4\cdot a \cdot c4⋅a

2

⋅x

2

+4⋅a⋅b⋅x=−4⋅a⋅c (Compatibility with multiplication)

4) 4\cdot a^{2}\cdot x^{2} + 4\cdot a \cdot b \cdot x + b^{2} = b^{2}-4\cdot a \cdot c4⋅a

2

⋅x

2

+4⋅a⋅b⋅x+b

2

=b

2

−4⋅a⋅c (Compatibility with addition/Existence of additive inverse/Modulative property)

5) (2\cdot a \cdot x + b)^{2} = b^{2}-4\cdot a \cdot c(2⋅a⋅x+b)

2

=b

2

−4⋅a⋅c (Perfect square trinomial)

According to steps 2 and 4. The second-order polynomial must be added by -c−c and b^{2}b

2

to create a perfect square trinomial.

for b its (t) and c its (d)

step-by-step explanation:

the answer is (-4,-4)

hope this ! : d