Question

How do you solve \[{\left( {x - 9} \right)^2} = 12\]?

Answer 1

Hint: We can solve this using algebraic identity \[{\left( {a - b} \right)^2} = {a^2} + {b^2} + 2ab\]. After applying this and simplifying we will have a quadratic equation. We can solve the obtained quadratic equation using factorization method or by quadratic formula or by completing the square. We will have two values for ‘x’.

Complete step-by-step solution:
Given, \[{\left( {x - 9} \right)^2} = 12\]
Now applying \[{\left( {a - b} \right)^2} = {a^2} + {b^2} + 2ab\], we have,
\[{x^2} + {9^2} - 18x = 12\]
\[{x^2} + 81 - 18x = 12\]
\[ \Rightarrow {x^2} - 18x + 69 = 0\]
On comparing the given equation with the standard quadratic equation \[a{x^2} + bx + c = 0\], we have\[a = 1\], \[b = - 18\] and \[c = 69\].
Here the factorization method files because the standard form of the factorization of quadratic equation is \[a{x^2} + {b_1}x + {b_2}x + c = 0\], which does not satisfies the condition \[{b_1} \times {b_2} = a \times c\] and \[{b_1} + {b_2} = b\].
Hence we use the quadratic formula, that is \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
On substituting we have,
 \[\Rightarrow x = \dfrac{{ - ( - 18) \pm \sqrt {{{( - 18)}^2} - 4(1)(69)} }}{{2(1)}}\]
\[\Rightarrow x = \dfrac{{18 \pm \sqrt {324 - 276} }}{2}\]
\[\Rightarrow x = \dfrac{{18 \pm \sqrt {48} }}{2}\]
We can write 48 as \[48 = 16 \times 4\],
\[\Rightarrow x = \dfrac{{18 \pm \sqrt {16 \times 3} }}{2}\]
We know that 16 is perfect square and we taking it outside the,
\[\Rightarrow x = \dfrac{{18 \pm 4\sqrt 3 }}{2}\]
Taking 2 common
\[\Rightarrow x = \dfrac{{2\left( {9 \pm 2\sqrt 3 } \right)}}{2}\]
\[ \Rightarrow x = 9 \pm 2\sqrt 3 \].

Thus we have two roots that is \[x = 9 + 2\sqrt 3 \]and \[x = 9 - 2\sqrt 3 \]


Note: We can also solve this easily.
We have \[{\left( {x - 9} \right)^2} = 12\]
Taking square root on both sides we have,
\[\Rightarrow\sqrt {{{\left( {x - 9} \right)}^2}} = \pm \sqrt {12} \]
\[\Rightarrow x - 9 = \pm \sqrt {12} \]
We can write 12 as product of 4 and 3,
\[\Rightarrow x - 9 = \pm \sqrt {4 \times 3} \]
\[\Rightarrow x - 9 = \pm 2\sqrt 3 \]
\[ \Rightarrow x = 9 \pm 2\sqrt 3 \]
Thus we have two roots, \[x = 9 + 2\sqrt 3 \] and \[x = 9 - 2\sqrt 3 \]. In both the cases we have the same answer.