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Please help me with this problem:Solve the quadratic equation x^2 + 2x = 35 using two different methodsAnswer:Method 1:Method 2:

Please help me with this problem:Solve the quadratic equation x^2 + 2x = 35 using-example-1
User Lovespeed
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Step-by-step explanation

We must solve by two different methods the following equation:


x^2+2x=35.

Method 1: Completing the square

1) We rewrite the equation above as:


x^2+2\cdot1\cdot x=35.

2) Now, we add 1² on both sides of the equation:


(x^2+2\cdot1\cdot x+1^2)=35+1^2=36.

3) We see that the left and right sides can be written as squares:


(x+1)^2=6^2.

4) Taking the square root on both sides, we get two solutions:


\begin{gathered} x+1=+6\Rightarrow x_1=6-1=5, \\ x+1=-6\Rightarrow x_2=-6-1=-7. \end{gathered}

Method 2: Using the quadratic formula

1) We rewrite the equation above as:


x^2+2x-35=0.

2) We identify a quadratic equation:


a\cdot x^2+b\cdot x+c=0.

With coefficients:

• a = 1,

,

• b = 2,

,

• c = -35.

The roots of this equation are given by:


\begin{gathered} x_1=(-b+√(b^2-4ac))/(2a), \\ x_2=(-b-√(b^2-4ac))/(2a). \end{gathered}

3) Replacing the coefficients of the quadratic equation in the formula above, we get:


\begin{gathered} x_1=(-2+√(2^2-4\cdot1\cdot(-35)))/(2\cdot1)=(-2+√(144))/(2)=(-2+12)/(2)=(10)/(2)=5, \\ x_2=(-2-√(2^2-4\cdot1\cdot(-35)))/(2\cdot1)=(-2-√(144))/(2)=(-2-12)/(2)=-(14)/(2)=-7. \end{gathered}Answer

The roots of the polynomial are:

• x₁ = 5

,

• x₂ = -7

User Martin Pfeffer
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