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Question: How do you solve the quadratic equation \({x^2} - 6x - 27 = 0\)?...

How do you solve the quadratic equation x2−6x−27=0{x^2} - 6x - 27 = 0?

Explanation

Solution

This problem deals with solving the roots of the given quadratic equation. The general and the standard form of the quadratic equation is considered as ax2+bx+c=0a{x^2} + bx + c = 0, where the roots or the solutions of this quadratic equation is given by the roots of the quadratic equation formula, which is given by as shown below:
⇒x=−b±b2−4ac2a\Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}

Complete step by step solution:
Consider the given quadratic equation as shown below:
⇒x2−6x−27=0\Rightarrow {x^2} - 6x - 27 = 0
Applying the formula of the roots of the quadratic equation formula to the above equation.
On comparing the above equation to the standard form of a quadratic equation which is generally given by ax2+bx+c=0a{x^2} + bx + c = 0.
Here on comparing the coefficients of the given quadratic equation to the general quadratic equation the coefficients are given by:
⇒a=1,b=−6\Rightarrow a = 1,b = - 6 and c=−27c = - 27.
Now applying the roots of the quadratic equation formula as shown below:
⇒x=−(−6)±(−6)2−4(1)(−27)2(1)\Rightarrow x = \dfrac{{ - \left( { - 6} \right) \pm \sqrt {{{\left( { - 6} \right)}^2} - 4\left( 1 \right)\left( { - 27} \right)} }}{{2\left( 1 \right)}}
⇒x=6±36+1082\Rightarrow x = \dfrac{{6 \pm \sqrt {36 + 108} }}{2}
On further simplifying the right hand side of the above equation as shown below:
⇒x=6±1442\Rightarrow x = \dfrac{{6 \pm \sqrt {144} }}{2}
We know that the square root of 144 is equal to 12, which is 144=12\sqrt {144} = 12, now substituting this in the above expression as shown below:
⇒x=6±122\Rightarrow x = \dfrac{{6 \pm 12}}{2}
Here xx takes two values, which are as shown below:
⇒x=6+122;x=6−122\Rightarrow x = \dfrac{{6 + 12}}{2};x = \dfrac{{6 - 12}}{2}
∴x=9;x=−3\therefore x = 9;x = - 3
The solutions of the equation x2−6x−27=0{x^2} - 6x - 27 = 0 are 9 and -3.

Note: Please note that the roots in the above solved quadratic equation are real and distinct. The nature of the roots of the quadratic equation can be determined by the discriminant of the quadratic equation which is given by:
⇒D=b2−4ac\Rightarrow D = {b^2} - 4ac
If D>0D > 0, the roots are real and distinct.
If D=0D = 0, the roots are real and equal.
If D<0D < 0, the roots are not real and complex.