How do you solve using the quadratic formula $6x + 9 = 2{x^2}$?
Answer
613.5k+ 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 $\dfrac{{3 \pm 3\sqrt 3 }}{2}$.
Recently Updated Pages
A Paragraph on Pollution in about 100-150 Words

What is BLO What is the full form of BLO class 8 social science CBSE

10 examples of friction in our daily life

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

XIX+XXX A 49 B 51 C 55 D 44 class 5 maths CBSE

Trending doubts
Explain the Treaty of Vienna of 1815 class 10 social science CBSE

1 GB equals how many MB?

10 examples of evaporation in daily life with explanations

What is the full form of POSCO class 10 social science CBSE

Which is the hottest planet in the Solar system A Earth class 10 social science CBSE

Name any four life processes in living things class 10 biology CBSE

