Question

Solve for \[x\] in \[{x^2} - 5x + 6 = 0\].

Answer 1

Hint: First factorise the given quadratic equation and then solve. Factorization is the process in which an expression is broken up into numbers which in turn can be multiplied to get the original expression.

Complete step by step solution:
For any quadratic equation, we always have two probable solutions for \[x\]. To factorize the above quadratic try to find any two numbers which when multiplied will be equal to the constant value \[ac\], and when added (in this case, or subtracted if the sign before the constant \[c\] is ‘\[ - \]’) will be equal to \[b\].
Here in this sum \[a = 1,\] \[b = 5\], \[c = 6\], \[ac = 6\] . Factorise \[6\] and try to find the required number. Observe that \[3\] when multiplied by \[2\] equals \[6\]and \[2\] when added to \[3\] gives \[5\] that is equal to \[b\].
Hence both the numbers are \[1\] and \[3\]. Now express \[b\] as the difference of the two numbers as shown:
\[\therefore \] \[{x^2} - 5x + 6\] \[ = \] \[0\]
\[ \Rightarrow \]\[{x^2} - (3x + 2x) + 6\] \[ = 0\]
\[ \Rightarrow {x^2} - 3x - 2x + 6\] \[ = 0\]
Take out the common factor from the first two terms and last two terms respectively:
\[ \Rightarrow \] \[x\left( {x - 3} \right) - 2\left( {x - 3} \right)\] \[ = \] \[0\]
Again take out the common factor:
\[ \Rightarrow \] \[\left( {x - 3} \right)\left( {x - 2} \right)\] \[ = \] \[0\]
Now, if \[ab = 0\], then either \[a = 0\] or \[b = 0\]:
\[\therefore \] \[\left( {x - 3} \right)\left( {x - 2} \right) = 0\]
\[ \Rightarrow \] either \[\left( {x - 3} \right) = 0\] or \[\left( {x - 2} \right) = 0\].

\[ \Rightarrow \] \[x = 3\] or \[x = 2\]

Additional information:
Any equation of the form \[a{x^2} + bx + c = 0\], \[a \ne 0\], where \[a,b,c\] are constants and \[x\] is a variable is known as a quadratic equation.
Here, \[{b^2} - 4ac\] is called the discriminant \[D\], of the quadratic equation.
For any quadratic equation:
If \[D > 0\], roots are real and unequal.
If \[D = 0\], roots are real and equal.
If \[D < 0\], roots are imaginary.

Note:
For any quadratic equation we always have two probable solutions for \[x\]. The above equation could also be solved by using the Sridharacharya method:
The formula for obtaining \[x\] in any quadratic equation of the form \[a{x^2} + bx + c = 0\], \[a \ne 0\], is given by:
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
In the given question \[a = 1\], \[b = - 5\], \[c = 6\]
\[\therefore \] By Sridharacharya method :
\[x = \dfrac{{ - ( - 5) \pm \sqrt {{{( - 5)}^2} - 4 \cdot 1 \cdot 6} }}{{2 \cdot 1}}\]
\[ \Rightarrow \] \[x = \dfrac{{5 \pm \sqrt {25 - 24} }}{2}\]
\[ \Rightarrow \] \[x = \dfrac{{5 \pm \sqrt 1 }}{2}\]
\[ \Rightarrow \] \[x = \dfrac{{5 + 1}}{2}\] or \[x = \dfrac{{5 - 1}}{2}\]
\[ \Rightarrow \] \[x = 3\] or, \[x = 2\].