Question

Solve the quadratic equation \[{x^2} + 12x + 35 = 0\].

Answer 1

Hint: We have a polynomial of degree 2 and it is called a quadratic equation. We can solve this using the factorization method or by the quadratic formula. Let us solve this using the quadratic formula. That is we use the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]. The quadratic formula is also called Sridhar’s formula.

Complete step-by-step solution:
Given, \[{x^2} + 12x + 35 = 0\].
Now consider the equation \[{x^2} + 12x + 35 = 0\]. 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 \[a = 1\], \[b = 12\] and \[c = 35\].
Now let us take the quadratic formula,
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
Substituting the values we have,
\[ \Rightarrow x = \dfrac{{ - 12 \pm \sqrt {{{12}^2} - 4(1)(35)} }}{{2(1)}}\]
\[ \Rightarrow x = \dfrac{{ - 12 \pm \sqrt {144 - 140} }}{2}\]
\[ \Rightarrow x = \dfrac{{ - 12 \pm \sqrt 4 }}{2}\]
We know that 4 is a perfect square,
\[ \Rightarrow x = \dfrac{{ - 12 \pm 2}}{2}\]
Thus we have two roots,
\[ \Rightarrow x = \dfrac{{ - 12 + 2}}{2}\] and \[x = \dfrac{{ - 12 - 2}}{2}\]
\[ \Rightarrow x = \dfrac{{ - 10}}{2}\] and \[x = \dfrac{{ - 14}}{2}\]
\[ \Rightarrow x = - 5\] and \[x = - 7\]. This is the required solution.
Additional information:
The highest exponent of the polynomial in a polynomial equation is called its degree. A polynomial equation has exactly as many roots as its degree. 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.

Note: The above equation can be solved using the factorization method. That is 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 \[{b_1} = 7\] and \[{b_2} = 5\]. Because \[{b_1} \times {b_2} = 35\] \[(a \times c)\] and \[{b_1} + {b_2} = 12(b)\]. If we split this and if we solve this we will have the same answer as above.