Question

Solve the quadratic equation. Write complex solutions in standard form. X2+35=5z

Answer 1

The quadratic is

`x^2+35=5x`

Subtract 5x from both sides

`\begin{gathered} x^2-5x+35=5x-5x \\ x^2-5x+35=0 \end{gathered}`

To solve it we will use the formula

`x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

a is the coefficient of x^2

b is the coefficient of x

c is the numerical term

Since a = 1, b = -5, c = 35

Then substitute them in the formula above

`x=\frac{-(-5)\pm\sqrt[]{(-5)^2-4(1)(35)}}{2(1)}`

Simplify it

`\begin{gathered} x=\frac{5\pm\sqrt[]{25-140}}{2} \\ x=\frac{5\pm\sqrt[]{-115}}{2} \end{gathered}`

since the square root of a -ve number is imaginary, then we will use i

`\begin{gathered} x=\frac{5\pm\sqrt[]{115}.\sqrt[]{-1}}{2} \\ x=\frac{5\pm\sqrt[]{115}i}{2} \\ \sqrt[]{-1}=i \end{gathered}`

The solutions are

`\mleft\lbrace\frac{5+\sqrt[]{115}i}{2},\frac{5-\sqrt[]{115}i}{2}\mright\rbrace`