Question

Solve: \[{x^2} - 6x + 5 = 0\]

Answer 1

Hint: In this given problem we need to factorize this polynomial expression. The given is a quadratic equation. We want to find the value of \[x\].The values of \[x\] can be find by using the quadratic formula: \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]. The polynomial equation whose highest degree is two is called a quadratic equation.

Complete step-by-step answer:
The given is a quadratic equation in the form\[a{x^2} + bx + c = 0\].
If no coefficient with variable then the variable have the value coefficient 1, i.e.,\[{x^2}\] means \[1{x^2}\]
Therefore the given equation is \[1{x^2} - 6x + 5 = 0\]
In the given question\[a = 1\] ,\[b = - 6\] ,\[c = 5\].
By using quadratic formula, \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Substitute values in the formula
\[x = \dfrac{{ - ( - 6) \pm \sqrt {{{( - 6)}^2} - 4(1)(5)} }}{{2(1)}}\]
    =\[\dfrac{{6 \pm \sqrt {36 - 20} }}{2}\]
    =\[\dfrac{{6 \pm \sqrt {16} }}{2}\]
Simplify the square root, we can get
    =\[\dfrac{{6 \pm 4}}{2}\]
( Discriminant\[{b^2} - 4ac > 0\], therefore there are two real values for\[x\].)
    =\[\dfrac{{6 + 4}}{2}\],\[\dfrac{{6 - 4}}{2}\]
    =\[\dfrac{{10}}{2}\],\[\dfrac{2}{2}\]
\[x = 5,1\]
The value of \[x = 5\]and \[x = 1\].

Note: Every quadratic equation is in the form \[a{x^2} + bx + c = 0\] where \[a,b,c\] are real numbers and\[a \ne 0\].
To know about the values first we should find the discriminant \[{b^2} - 4ac\].
If \[{b^2} - 4ac > 0\], there are two real roots (values for\[x\]).
If \[{b^2} - 4ac < 0\], there are two complex roots.
In quadratic equations, imaginary numbers (and complex roots) occur when the value under the radical portion of the quadratic formula is negative (\[{b^2} - 4ac < 0\]). When this occurs, the equation has no roots in the set of real numbers. The roots belong to the set of complex numbers, and will be called "complex roots" (or "imaginary roots"). These complex roots will be expressed in the form \[a \pm bi\].
If \[{b^2} - 4ac = 0\]there is one real root.
If any term is missing in the quadratic equation then consider it as 0.eg: \[{x^2} - 16 = 0\].
Here the quadratic equation in the form \[a{x^2} - c = 0\] where \[bx\] is missing let consider \[bx\] as \[0\]then the given quadratic equation is \[{x^2} - 0x - 16 = 0\]. \[0\] has no value now we can substitute \[0\]in the quadratic formula to find the roots.