How do you solve the quadratic equation \[ - {x^2} + 5x - 9 = 0\] ?
Answer
579.6k+ views
Hint: Use method of determinant to solve for the value of x from the given quadratic equation. Compare the quadratic equation with general quadratic equation and substitute values in the formula of finding roots of the equation. Solve the value under the square root and write two roots by separating plus and minus signs from the obtained answer.
* For a general quadratic equation \[a{x^2} + bx + c = 0\], roots are given by formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Complete step by step solution:
We are given the quadratic equation \[ - {x^2} + 5x - 9 = 0\] … (1)
We know that the general quadratic equation is \[a{x^2} + bx + c = 0\] where ‘a’, ‘b’, and ‘c’ are constant values.
On comparing the quadratic equation in equation (1) with general quadratic equation\[a{x^2} + bx + c = 0\], we get \[a = - 1,b = 5,c = - 9\]
Substitute the values of a, b and c in the formula of finding roots of the equation.
\[ \Rightarrow x = \dfrac{{ - (5) \pm \sqrt {{{(5)}^2} - 4 \times ( - 1) \times ( - 9)} }}{{2 \times ( - 1)}}\]
Square the terms under the square root in numerator of the fraction
\[ \Rightarrow x = \dfrac{{ - 5 \pm \sqrt {25 - 4 \times (9)} }}{{2 \times ( - 1)}}\]
Multiply the values inside the square root in numerator of the fraction
\[ \Rightarrow x = \dfrac{{ - 5 \pm \sqrt {25 - 36} }}{{ - 2}}\]
\[ \Rightarrow x = \dfrac{{ - 5 \pm \sqrt { - 11} }}{{ - 2}}\]
We know that \[\sqrt { - 1} = i\]
\[ \Rightarrow x = \dfrac{{ - 5 \pm i\sqrt {11} }}{{ - 2}}\]
So, complex roots of the quadratic equation are:
\[x = \dfrac{{ - 5 - i\sqrt {11} }}{{ - 2}}\]and \[x = \dfrac{{ - 5 + i\sqrt {11} }}{{ - 2}}\]
i.e. \[x = \dfrac{{5 + i\sqrt {11} }}{2}\]and \[x = \dfrac{{5 - i\sqrt {11} }}{2}\]
\[\therefore \] Solution of the equation \[ - {x^2} + 5x - 9 = 0\] is \[x = \dfrac{{5 + i\sqrt {11} }}{2}\]and \[x = \dfrac{{5 - i\sqrt {11} }}{2}\].
Note: Many students leave their answer in fraction form, or without cancelling common factors which is okay as the answer can be according to the requirement of the question, but keep in mind we should always solve the fractions to the simplest form.
Many students make the mistake of solving for the roots or values of x forcefully for this question using factorization method which is wrong. Many students leave their answer with a negative term under the square root which is wrong as we clearly know the value of the negative term under the square root means the presence of ‘i’ i.e. the number becomes a complex number.
* For a general quadratic equation \[a{x^2} + bx + c = 0\], roots are given by formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Complete step by step solution:
We are given the quadratic equation \[ - {x^2} + 5x - 9 = 0\] … (1)
We know that the general quadratic equation is \[a{x^2} + bx + c = 0\] where ‘a’, ‘b’, and ‘c’ are constant values.
On comparing the quadratic equation in equation (1) with general quadratic equation\[a{x^2} + bx + c = 0\], we get \[a = - 1,b = 5,c = - 9\]
Substitute the values of a, b and c in the formula of finding roots of the equation.
\[ \Rightarrow x = \dfrac{{ - (5) \pm \sqrt {{{(5)}^2} - 4 \times ( - 1) \times ( - 9)} }}{{2 \times ( - 1)}}\]
Square the terms under the square root in numerator of the fraction
\[ \Rightarrow x = \dfrac{{ - 5 \pm \sqrt {25 - 4 \times (9)} }}{{2 \times ( - 1)}}\]
Multiply the values inside the square root in numerator of the fraction
\[ \Rightarrow x = \dfrac{{ - 5 \pm \sqrt {25 - 36} }}{{ - 2}}\]
\[ \Rightarrow x = \dfrac{{ - 5 \pm \sqrt { - 11} }}{{ - 2}}\]
We know that \[\sqrt { - 1} = i\]
\[ \Rightarrow x = \dfrac{{ - 5 \pm i\sqrt {11} }}{{ - 2}}\]
So, complex roots of the quadratic equation are:
\[x = \dfrac{{ - 5 - i\sqrt {11} }}{{ - 2}}\]and \[x = \dfrac{{ - 5 + i\sqrt {11} }}{{ - 2}}\]
i.e. \[x = \dfrac{{5 + i\sqrt {11} }}{2}\]and \[x = \dfrac{{5 - i\sqrt {11} }}{2}\]
\[\therefore \] Solution of the equation \[ - {x^2} + 5x - 9 = 0\] is \[x = \dfrac{{5 + i\sqrt {11} }}{2}\]and \[x = \dfrac{{5 - i\sqrt {11} }}{2}\].
Note: Many students leave their answer in fraction form, or without cancelling common factors which is okay as the answer can be according to the requirement of the question, but keep in mind we should always solve the fractions to the simplest form.
Many students make the mistake of solving for the roots or values of x forcefully for this question using factorization method which is wrong. Many students leave their answer with a negative term under the square root which is wrong as we clearly know the value of the negative term under the square root means the presence of ‘i’ i.e. the number becomes a complex number.
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