Question

How do you solve \[{\left( {x - 7} \right)^2} = 9\]?

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, \[{\left( {x - 7} \right)^2} = 9\]
Now applying \[{\left( {a - b} \right)^2} = {a^2} + {b^2} + 2ab\], we have,
\[\Rightarrow {x^2} + {7^2} - 14x = 9\]
\[\Rightarrow {x^2} + 49 - 14x = 9\]
\[\Rightarrow {x^2} + 49 - 14x - 9 = 0\]
\[ \Rightarrow {x^2} - 14x + 40 = 0\]
On comparing the given equation with the standard quadratic equation \[a{x^2} + bx + c = 0\], we have \[a = 1\], \[b = - 14\] and \[c = 40\].
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 \[{b_1} + {b_2} = b\].
That is \[{x^2} - 10x - 4x + 40 = 0\] which does not satisfies the condition \[{b_1} \times {b_2} = a \times c\] and \[{b_1} + {b_2} = b\].
Hence, we have,
\[\Rightarrow {x^2} - 10x - 4x + 40 = 0\]
Taking x common in the first two terms and -4 common in the remaining two terms,
\[\Rightarrow x(x - 10) - 4(x - 10) = 0\]
Take \[(x - 10)\] as common,
\[\Rightarrow (x - 10)(x - 4) = 0\]
By zero product principle we have,
\[\Rightarrow x - 10 = 0\] and \[x - 4 = 0\]
\[ \Rightarrow x = 10\] and \[x = 4\]

Thus the values of x are x = 10 and x = 4.

Note: We can also solve this easily.
We have \[{\left( {x - 7} \right)^2} = 9\]
Taking square root on both sides we have,
\[\Rightarrow \sqrt {{{\left( {x - 7} \right)}^2}} = \pm \sqrt 9 \]
\[\Rightarrow x - 7 = \pm \sqrt 9 \]
We know that 9 is a perfect square,
\[\Rightarrow x - 7 = \pm 3\]
That is,
\[\Rightarrow x - 7 = 3\] and \[x - 7 = - 3\]
\[ \Rightarrow x = 3 + 7\] and \[x = - 3 + 7\]
\[ \Rightarrow x = 10\] and \[x = 4\]
In both the cases we have the same answer.