Question

Solve the given quadratic equation \[{\left( {x + 8} \right)^2} - 5 = 31\].

Answer 1

Hint: We can solve this in two methods. First we expand the terms in the brackets using the algebraic identity and then simplifying we will have a quadratic equation. Quadratic equation, we can solve this using factorization method or by quadratic formula or by graphing method or by completing square method. We solve this using the factorization method. If we fail to split the middle term we use the quadratic formula, that is \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].

Complete step-by-step solution:
Given, \[{\left( {x + 8} \right)^2} - 5 = 31\].
We know \[{(a + b)^2} = {a^2} + {b^2} + 2ab\]
Applying this we have
\[{x^2} + {8^2} + 2(x)(8) - 5 = 31\]
\[ \Rightarrow {x^2} + 64 + 16x - 5 - 31 = 0\]
\[ \Rightarrow {x^2} + 16x + 64 - 5 - 31 = 0\]
\[ \Rightarrow {x^2} + 16x + 28 = 0\]
Now consider the equation \[{x^2} + 16x + 28 = 0\]. The degree of this equation is two hence we will have two factors.
On comparing the given equation with the standard quadratic equation \[a{x^2} + bx + c = 0\]. We have \[a = 1\], \[b = 16\] and \[c = 28\].
For factorization, the standard equation is rewritten as \[a{x^2} + {b_1}x + {b_2}x + c = 0\] such that \[{b_1} \times {b_2} = ac\] and \[{b_1} + {b_2} = b\].
Here we can say that \[{b_1} = 14\] and \[{b_2} = 2\]. Because \[{b_1} \times {b_2} = 28\] \[(a \times c)\] and \[{b_1} + {b_2} = 16(b)\].
Now we write \[{x^2} + 16x + 28 = 0\] as,
\[ \Rightarrow {x^2} + 14x + 2x + 28 = 0\]
Taking ‘x’ common in the first two terms and taking $2$ common in the remaining two terms we have,
\[ \Rightarrow x(x + 14) + 2(x + 14) = 0\]
Again taking \[(x + 14)\] common we have,
\[(x + 14)(x + 2) = 0\].
By zero product principle we have,
 \[ \Rightarrow x + 14 = 0\] and \[x + 2 = 0\]
\[ \Rightarrow x = - 14\] and \[x = - 2\]
This is the required solution.

Note: We can solve this using another method.
 \[{\left( {x + 8} \right)^2} - 5 = 31\]
Shifting -5 on the right hand side of the equation we have,
\[{\left( {x + 8} \right)^2} = 31 + 5\]
\[{\left( {x + 8} \right)^2} = 36\]
taking square root on both sides we have
\[\sqrt {{{\left( {x + 8} \right)}^2}} = \pm \sqrt {36} \]
\[x + 8 = \pm 6\]
that is we have
\[ \Rightarrow x + 8 = 6\] and \[ \Rightarrow x + 8 = - 6\]
\[ \Rightarrow x = 6 - 8\] and \[ \Rightarrow x = - 6 - 8\]
\[ \Rightarrow x = - 14\] and \[x = - 2\]
In both cases we have the same answer.