X¹,²=+7
You could use several methods.
The fastest is:
x2=49
∴√x2=±√49
x1,2=±7
I'd prefer to factorize the equation using the Square Difference Identitiy:
(a2−b2)=(a+b)(a−b)
to rewrite the second degree equation as product of first degree equation.
In your exercise:
(x2−49)=0⇔(x2−72)=0⇔(x+7)(x−7)=0
A product is zero when the factors are zero
∴(x+7)=0⇒x1=−7
∴(x−7)=0⇒x2=7
Therefore:
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Answers & Comments
X¹,²=+7
You could use several methods.
The fastest is:
x2=49
∴√x2=±√49
x1,2=±7
I'd prefer to factorize the equation using the Square Difference Identitiy:
(a2−b2)=(a+b)(a−b)
to rewrite the second degree equation as product of first degree equation.
In your exercise:
(x2−49)=0⇔(x2−72)=0⇔(x+7)(x−7)=0
A product is zero when the factors are zero
∴(x+7)=0⇒x1=−7
∴(x−7)=0⇒x2=7
Therefore:
x1,2=±7