Question

How do you solve \[x(x + 6) = - 15\] using the quadratic formula?

Answer 1

Hint: If we have a polynomial of degree ‘n’ then we have ‘n’ number of roots or factors. After simplifying the given problem we will have a quadratic equation. A polynomial of degree two is called a quadratic polynomial and its zeros can be found using many methods like factorization, completing the square, graphs, quadratic formula etc. The quadratic formula is used when we fail to find the factors of the equation. In the given question we have to solve the given quadratic equation using the quadratic formula. That is \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] .

Complete step by step solution:
Given, \[x(x + 6) = - 15\] .
Simplifying we have,
 \[{x^2} + 6x = - 15\]
 \[{x^2} + 6x + 15 = 0\]
On comparing the given equation with the standard quadratic equation \[a{x^2} + bx + c = 0\] . We get, \[a = 1\] , \[b = 6\] and \[c = 15\] .
Substituting in the formula of standard quadratic, \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] .
 \[ \Rightarrow x = \dfrac{{ - (6) \pm \sqrt {{{(6)}^2} - 4(1)(15)} }}{{2(1)}}\]
 \[ \Rightarrow x = \dfrac{{ - 6 \pm \sqrt {36 - 60} }}{2}\]
 \[ \Rightarrow x = \dfrac{{ - 6 \pm \sqrt { - 24} }}{2}\]
We know that \[\sqrt { - 1} = i\] ,
 \[ \Rightarrow x = \dfrac{{ - 6 \pm i\sqrt {24} }}{2}\]
We can write 24 as \[24 = 4 \times 6\]
 \[ \Rightarrow x = \dfrac{{ - 6 \pm i\sqrt {4 \times 6} }}{2}\]
We know that 4 is a perfect square and taking it outside we have,
 \[ \Rightarrow x = \dfrac{{ - 6 \pm 2i\sqrt 6 }}{2}\]
Taking 2 common we have,
 \[ \Rightarrow x = \dfrac{{2\left( { - 3 \pm i\sqrt 6 } \right)}}{2}\]
 \[ \Rightarrow x = - 3 \pm i\sqrt 6 \]
Thus we have two roots,
 \[ \Rightarrow x = - 3 + i\sqrt 6 \] and \[x = - 3 - i\sqrt 6 \]
So, the correct answer is “ \[ x = - 3 + i\sqrt 6 \] and \[x = - 3 - i\sqrt 6 \] ”.

Note: In various fields of mathematics require the point at which the value of a polynomial is zero, those values are called the factors/solution/zeros of the given polynomial. On the x-axis, the value of y is zero so the roots of an equation are the points on the x-axis that is the roots are simply the x-intercepts. We cannot solve this by simple factorization. That is by expanding the middle term into a sum of two terms, such that the product of two terms is equal to the product of ‘a’ and ‘c’, sum of two terms is equal to ‘b’. We also know that the quadratic formula is also known as Sridhar’s formula.