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Question

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

Solve the quadratic equation x2+12x+35=0{x^2} + 12x + 35 = 0.

Explanation

Solution

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=b±b24ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}. The quadratic formula is also called Sridhar’s formula.

Complete step-by-step solution:
Given, x2+12x+35=0{x^2} + 12x + 35 = 0.
Now consider the equation x2+12x+35=0{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 ax2+bx+c=0a{x^2} + bx + c = 0. We have a=1a = 1, b=12b = 12 and c=35c = 35.
Now let us take the quadratic formula,
x=b±b24ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}.
Substituting the values we have,
x=12±1224(1)(35)2(1)\Rightarrow x = \dfrac{{ - 12 \pm \sqrt {{{12}^2} - 4(1)(35)} }}{{2(1)}}
x=12±1441402\Rightarrow x = \dfrac{{ - 12 \pm \sqrt {144 - 140} }}{2}
x=12±42\Rightarrow x = \dfrac{{ - 12 \pm \sqrt 4 }}{2}
We know that 4 is a perfect square,
x=12±22\Rightarrow x = \dfrac{{ - 12 \pm 2}}{2}
Thus we have two roots,
x=12+22\Rightarrow x = \dfrac{{ - 12 + 2}}{2} and x=1222x = \dfrac{{ - 12 - 2}}{2}
x=102\Rightarrow x = \dfrac{{ - 10}}{2} and x=142x = \dfrac{{ - 14}}{2}
x=5\Rightarrow x = - 5 and x=7x = - 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 ax2+b1x+b2x+c=0a{x^2} + {b_1}x + {b_2}x + c = 0 such thatb1×b2=ac{b_1} \times {b_2} = ac andb1+b2=b{b_1} + {b_2} = b.
Here we can say that b1=7{b_1} = 7 and b2=5{b_2} = 5. Because b1×b2=35{b_1} \times {b_2} = 35 (a×c)(a \times c) and b1+b2=12(b){b_1} + {b_2} = 12(b). If we split this and if we solve this we will have the same answer as above.