Question

Hello. solve for x in the equation x2 - 10x+25=35?

Answer 1

`\begin{gathered} \text{Given} \\ x^2-10x+25=35 \end{gathered}`

Move the constant on the right side to the left side by subtracting both sides by 35

`\begin{gathered} x^2-10x+25=35 \\ x^2-10x+25-35=35-35 \\ x^2-10x-10=\cancel{0} \end{gathered}`

Now that it is in the standard form of quadratic equation, we can use the quadratic formula to solve for x

`\begin{gathered} \text{The standard form is }ax^2+bx+c=0 \\ \\ \text{In }x^2-10x-10=0, \\ a=1,b=-10,c=-10 \end{gathered}`
`\begin{gathered} \text{Use the quadratic formula} \\ x=( -b \pm√(b^2 - 4ac))/( 2a ) \\ \\ \text{Substitute the values for }a,b\text{ and }c \\ x=( -(-10) \pm√((-10)^2 - 4(1)(-10)))/( 2(1) ) \\ x=\frac{10\pm\sqrt[]{100-(-40)}}{2} \\ x=\frac{10\pm\sqrt[]{100+40}}{2} \\ x=( 10 \pm√(140))/( 2 ) \\ x=( 10 \pm2√(35)\, )/( 2 ) \\ x=( 10 )/( 2 )\pm(2√(35)\, )/( 2 ) \\ x=5\pm\frac{\cancel{2}\sqrt[]{35}}{\cancel{2}} \\ x=5\pm\sqrt[]{35} \end{gathered}`

Therefore the solution for the given equation is

`\begin{gathered} x=5+\sqrt[]{35}\text{ or }x=10.9161 \\ \text{AND} \\ x=5-\sqrt[]{35}\text{ or }x=-0.9161 \end{gathered}`