
Solve the given quadratic equation:- \[2{x^2} + x - 6 = 0\]
Answer
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Hint: We have a given quadratic equation go to the and we have to find its root. So firstly we compare the equation with the general quadratic equation \[a{x^2} + bx + c = 0\] the value of \[A,B,C\]. Then we calculate the discriminants of the equation the value of discriminant tells us about nature off the root. If \[D > 0\]. Roots are real and distinct. If \[D = 0\]Roots are real. If \[D < 0\] Roots are not real. After finding the natural root. We apply the quadratic formula method which is given as.
\[x = \dfrac{{ - b \pm \sqrt D }}{{2a}}\]Here\[\;D\] is discriminant will given two values of \[x\] which are of the root of the equation.
Complete step-by-step answer:
The given quadratic equation is \[2{x^2} + x - 6 = 0\].We have to find its roots.
Comparing the equation with \[a{x^2} + bx + c = 0\]
\[a = 2{\text{ }}b = 1,c = - 6\]
Now we calculate the discriminant.
Discriminant will tell us the nature of root
\[ \Rightarrow D = {b^2} - 4ac\]
\[ \Rightarrow 1 - 4 \times 2 \times ( - 6) = 1 + 48 = 49\]
\[ \Rightarrow D = 49 > 0\]
So the root of the given equation are root and distinct
Now \[x = \dfrac{{ - b \pm \sqrt D }}{{2a}} = \dfrac{{ - 1 \pm \sqrt {49} }}{{2 \times 2}} = \dfrac{{ - 1 \pm 7}}{4}\]
We can write the roots separately ,
\[ \Rightarrow x = \dfrac{{ - 1 + 7}}{4},\] \[x = \dfrac{{ - 1 - 7}}{4}\]
By rearranging, We get this values
\[ \Rightarrow x = \dfrac{6}{4}\], \[x = \dfrac{{ - 8}}{4}\]
By rearranging, We get this two values of \[x\] that are given below:
\[ \Rightarrow x = \dfrac{3}{2},\] \[x = - 2\]
So the roots of the equation are \[2{x^2} + x - 6 = 0\] are \[\dfrac{3}{2}\] and \[ - 2\].
Note: The polynomial equation which can be written in \[ax + bc + c = 0\] is called a quadratic equation. The degree of a quadratic equation is 2. Here x is an unknown variable \[a,b,c,\]and \[c\] are constants or called the coefficient of x. And x. Roots of quadratic equations are those values of \[c\] which gives the value of the equation equal to zero called root of x .The roots of the quadratic equation may be real and distinct, real and equal, and no real root.
\[x = \dfrac{{ - b \pm \sqrt D }}{{2a}}\]Here\[\;D\] is discriminant will given two values of \[x\] which are of the root of the equation.
Complete step-by-step answer:
The given quadratic equation is \[2{x^2} + x - 6 = 0\].We have to find its roots.
Comparing the equation with \[a{x^2} + bx + c = 0\]
\[a = 2{\text{ }}b = 1,c = - 6\]
Now we calculate the discriminant.
Discriminant will tell us the nature of root
\[ \Rightarrow D = {b^2} - 4ac\]
\[ \Rightarrow 1 - 4 \times 2 \times ( - 6) = 1 + 48 = 49\]
\[ \Rightarrow D = 49 > 0\]
So the root of the given equation are root and distinct
Now \[x = \dfrac{{ - b \pm \sqrt D }}{{2a}} = \dfrac{{ - 1 \pm \sqrt {49} }}{{2 \times 2}} = \dfrac{{ - 1 \pm 7}}{4}\]
We can write the roots separately ,
\[ \Rightarrow x = \dfrac{{ - 1 + 7}}{4},\] \[x = \dfrac{{ - 1 - 7}}{4}\]
By rearranging, We get this values
\[ \Rightarrow x = \dfrac{6}{4}\], \[x = \dfrac{{ - 8}}{4}\]
By rearranging, We get this two values of \[x\] that are given below:
\[ \Rightarrow x = \dfrac{3}{2},\] \[x = - 2\]
So the roots of the equation are \[2{x^2} + x - 6 = 0\] are \[\dfrac{3}{2}\] and \[ - 2\].
Note: The polynomial equation which can be written in \[ax + bc + c = 0\] is called a quadratic equation. The degree of a quadratic equation is 2. Here x is an unknown variable \[a,b,c,\]and \[c\] are constants or called the coefficient of x. And x. Roots of quadratic equations are those values of \[c\] which gives the value of the equation equal to zero called root of x .The roots of the quadratic equation may be real and distinct, real and equal, and no real root.
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