Question

X²–5x+5=0what is the true solution of this equation?

Answer 1

Here, we are to give the true solution of this equation.

To proceed with solving a quadratic equation, we need to use the factorization method easily by checking out for factors.

In events where there are no possible factors or it is not obviously factorisable, we proceed to use the quadratic formula.

The quadratic formula is given below;

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

a refers to the coefficient of x^2 which is 1 in this case

b is the coefficient of x which is -5 in this case

c is the last number which is 5 in this case

To solve the quadratic equation, we insert these value into the quadratic equation formula;

`\begin{gathered} x\text{ = }\frac{-(-5)\pm\text{ }\sqrt[]{(-5)^2\text{ - 4(1)(5)}}}{2(1)} \\ \\ x\text{ = }\frac{5\text{ }\pm\text{ }\sqrt[]{25-20}}{2} \\ \\ x\text{ = }\frac{5\text{ }\pm\sqrt[]{5}}{2} \\ \\ x\text{ = }\frac{5\text{ + }\sqrt[]{5}}{2}\text{ or }\frac{5-\sqrt[]{5}}{2} \end{gathered}`

The two values of x above are the true solution to this equation