
How do you solve using the quadratic formula $6x + 9 = 2{x^2}$?
Answer
510.3k+ views
Hint: We will first of all, write the general quadratic equation and the formula for its roots and then on comparing put the values as in formula and thus we have the required roots.
Complete step by step solution:
We are given that we are required to solve $6x + 9 = 2{x^2}$ using the quadratic formula.
We can write this equation as: $2{x^2} - 6x - 9 = 0$
The general quadratic equation is given by $a{x^2} + bx + c = 0$, where a, b and c are the constants.
The roots are this equation is given by the following expression with us:-
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Comparing the general quadratic equation, we have: a = 2, b = - 6 and c = - 9.
Therefore, the roots of the equation $2{x^2} - 6x - 9 = 0$ are given by:-
$ \Rightarrow x = \dfrac{{ - ( - 6) \pm \sqrt {{{( - 6)}^2} - 4(2)( - 9)} }}{{2(2)}}$
Simplifying the calculations, we get the following equation with us:-
$ \Rightarrow x = \dfrac{{6 \pm \sqrt {36 + 72} }}{4}$
Simplifying the calculations further, we get the following equation with us:-
$ \Rightarrow x = \dfrac{{6 \pm 6\sqrt 3 }}{4}$
Crossing – off 2 from both the numerator and denominator, we will get:-
The possible values of x as $\dfrac{{3 \pm 3\sqrt 3 }}{2}$.
Note:
The students must note that there is an alternate way to do the same question, it is given as follows:-
We are given that we are required to solve $6x + 9 = 2{x^2}$ using the quadratic formula.
We can write this equation as: ${x^2} - 3x - \dfrac{9}{2} = 0$.
Here, we will use the method of completing the square to find the roots of the given equation.
We know we can write the above equation as ${x^2} - 2\left( {\dfrac{3}{2}} \right)x + {\left( {\dfrac{3}{2}} \right)^2} - {\left( {\dfrac{3}{2}} \right)^2} - \dfrac{9}{2} = 0$
Since we know that we have an identity given by the expression: ${(a - b)^2} = {a^2} + {b^2} - 2ab$
We can further write the above expression ${x^2} - 2\left( {\dfrac{3}{2}} \right)x + {\left( {\dfrac{3}{2}} \right)^2} - {\left( {\dfrac{3}{2}} \right)^2} - \dfrac{9}{2} = 0$ as: ${\left( {x - \dfrac{3}{2}} \right)^2} - {\left( {\dfrac{3}{2}} \right)^2} - \dfrac{9}{2} = 0$
Now, simplifying it, we will then obtain the following equation with us:-
$ \Rightarrow {\left( {x - \dfrac{3}{2}} \right)^2} - \dfrac{9}{4} - \dfrac{9}{2} = 0$
Simplifying it further, we will then obtain the following equation with us:-
$ \Rightarrow {\left( {x - \dfrac{3}{2}} \right)^2} = \dfrac{{27}}{4}$
Taking square – root on both the sides, we will then obtain the following equation:-
\[ \Rightarrow x - \dfrac{3}{2} = \dfrac{{3\sqrt 3 }}{2}\]
Hence, the possible values of x as $\dfrac{{3 \pm 3\sqrt 3 }}{2}$.
Complete step by step solution:
We are given that we are required to solve $6x + 9 = 2{x^2}$ using the quadratic formula.
We can write this equation as: $2{x^2} - 6x - 9 = 0$
The general quadratic equation is given by $a{x^2} + bx + c = 0$, where a, b and c are the constants.
The roots are this equation is given by the following expression with us:-
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Comparing the general quadratic equation, we have: a = 2, b = - 6 and c = - 9.
Therefore, the roots of the equation $2{x^2} - 6x - 9 = 0$ are given by:-
$ \Rightarrow x = \dfrac{{ - ( - 6) \pm \sqrt {{{( - 6)}^2} - 4(2)( - 9)} }}{{2(2)}}$
Simplifying the calculations, we get the following equation with us:-
$ \Rightarrow x = \dfrac{{6 \pm \sqrt {36 + 72} }}{4}$
Simplifying the calculations further, we get the following equation with us:-
$ \Rightarrow x = \dfrac{{6 \pm 6\sqrt 3 }}{4}$
Crossing – off 2 from both the numerator and denominator, we will get:-
The possible values of x as $\dfrac{{3 \pm 3\sqrt 3 }}{2}$.
Note:
The students must note that there is an alternate way to do the same question, it is given as follows:-
We are given that we are required to solve $6x + 9 = 2{x^2}$ using the quadratic formula.
We can write this equation as: ${x^2} - 3x - \dfrac{9}{2} = 0$.
Here, we will use the method of completing the square to find the roots of the given equation.
We know we can write the above equation as ${x^2} - 2\left( {\dfrac{3}{2}} \right)x + {\left( {\dfrac{3}{2}} \right)^2} - {\left( {\dfrac{3}{2}} \right)^2} - \dfrac{9}{2} = 0$
Since we know that we have an identity given by the expression: ${(a - b)^2} = {a^2} + {b^2} - 2ab$
We can further write the above expression ${x^2} - 2\left( {\dfrac{3}{2}} \right)x + {\left( {\dfrac{3}{2}} \right)^2} - {\left( {\dfrac{3}{2}} \right)^2} - \dfrac{9}{2} = 0$ as: ${\left( {x - \dfrac{3}{2}} \right)^2} - {\left( {\dfrac{3}{2}} \right)^2} - \dfrac{9}{2} = 0$
Now, simplifying it, we will then obtain the following equation with us:-
$ \Rightarrow {\left( {x - \dfrac{3}{2}} \right)^2} - \dfrac{9}{4} - \dfrac{9}{2} = 0$
Simplifying it further, we will then obtain the following equation with us:-
$ \Rightarrow {\left( {x - \dfrac{3}{2}} \right)^2} = \dfrac{{27}}{4}$
Taking square – root on both the sides, we will then obtain the following equation:-
\[ \Rightarrow x - \dfrac{3}{2} = \dfrac{{3\sqrt 3 }}{2}\]
Hence, the possible values of x as $\dfrac{{3 \pm 3\sqrt 3 }}{2}$.
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