Question

Solve for \[x\] in \[{x^2} - 25 = 0\].

Answer 1

Hint:First factorise the given quadratic equation and then solve. 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.

Complete step by step solution:
The given quadratic equation is : \[{x^2} - 25 = 0\].
First, factorize the equation:
Recall the formula \[{a^2} - {b^2}\] \[ = \] \[\left( {a + b} \right)\left( {a - b} \right)\].
Note that the given equation is also of the form \[{a^2} - {b^2}\] \[ = \] \[0\], where \[a = x\] and \[b = 5\], therefore applying the above formula:
\[{x^2} - 25 = 0\]
\[ \Rightarrow \] \[{x^2} - {(5)^2} = 0\]
\[ \Rightarrow \] \[\left( {x + 5} \right)\left( {x - 5} \right) = 0\]
Now, if \[ab = 0\], then either \[a = 0\] or \[b = 0\]:
\[\therefore \] \[\left( {x + 5} \right)\left( {x - 5} \right) = 0\]
\[ \Rightarrow \] either \[\left( {x + 5} \right) = 0\] or \[\left( {x - 5} \right) = 0\].
\[ \Rightarrow \] \[x = - 5\] or \[x = 5\]

\[\therefore x = \pm 5\].

Additional information:
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 \[a = 1\], \[b = 0\] (as there is no term involving \[x\]), \[c = - 25\]
\[\therefore \] By sridharacharya method :
\[x = \dfrac{{ - 0 \pm \sqrt {{0^2} - 4 \cdot 1 \cdot \left( { - 25} \right)} }}{{2 \cdot 1}}\]
\[ \Rightarrow \] \[x = \dfrac{{0 \pm \sqrt {0 + 100} }}{2}\] [since \[( - ) \times ( - ) = ( + )\]].
\[ \Rightarrow \] \[x = \pm \dfrac{{10}}{2}\] [since \[\sqrt {100} \] \[ = \] \[10\]].
\[ \Rightarrow \] \[x = \pm 5\].