Question

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

Answer 1

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:

$\begin{gathered} x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}, \\ x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}. \end{gathered}$

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

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

The roots of the polynomial are:

• x₁ = 5

,

• x₂ = -7