
Solve the following quadratic equation by the formula method : $3{x^2} + 7x + 2 = 0$.
Answer
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Hint: We have given a quadratic equation and find its solution by quadratic formula method. We first compare its quadratic equation with the standard quadratic equation $a{x^2} + bx + c = 0$ and find the value of $a,b$ and $c$ . Then firstly we check the nature of the root. This can be discriminant. If $D$ is greater than zero root will be real and distinct. If $D = 0$ then root will be real and equal. If $D < 0$ the real root does not exist. Then we apply a quadratic formula.
$x = \dfrac{{b + \sqrt D }}{{2a}}$
This will give us two roots of the equation.
Complete step by step answer:
We have given a quadratic equation and we have found its roots by quadratic formula method.
The quadratic equation is $3{x^2} + 7x + 2 = 0$
We compare this equation with $a{x^2} + bx + c = 0$
$a{x^2} + bx + c = 0$ is a general quadratic equation $a = 3,b = 7,c = 2$
We check the nature of the root first.
We find the discriminant for this
$D = {b^2} - 4ac$
If $D > 0$, then roots are real and distinct.
If $D = 0$, then roots are real and equal.
If $D < 0$, then roots are imaginary.
Now $D = {\left( 7 \right)^2} - 4 \times 3 \times 2$
$ = 49 - 24$
$D = 25 > 0$
$D > 0$, so roots are real and distinct.
The quadratic formula is given as
$x = \dfrac{{b \pm \sqrt D }}{{2a}}$, $D$ is discriminant
So $x = \dfrac{{ - 7 \pm \sqrt {25} }}{{2 \times 3}}$ \[ = \dfrac{{ - 7 \pm 5}}{6}\]
$x = \dfrac{{ - 7 \pm 5}}{6}$
$x = \dfrac{{ - 7 + 5}}{6},\dfrac{{ - 7 - 5}}{6}$
$x = \dfrac{{ - 2}}{6},\dfrac{{ - 12}}{6}$
$x = \dfrac{{ - 1}}{3}, - 3$
So, roots of the quadratic equation $3{x^2} + 7x + 2 = 0$ is $\dfrac{{ - 1}}{3}, - 3$
Note: Quadratic Equation : A quadratic equation is any equation that can be rearranged in standard form as there it represents unknown and a, b, c represent known number. Where $a \ne D$. If $a = 0$ then it will not be a quadratic equation.
The quadratic equation is the equation of parabola. There are two solutions of the quadratic equation; these solutions are called roots equations and these solutions are called roots. The roots may be real distinct, real and same and not real.
$x = \dfrac{{b + \sqrt D }}{{2a}}$
This will give us two roots of the equation.
Complete step by step answer:
We have given a quadratic equation and we have found its roots by quadratic formula method.
The quadratic equation is $3{x^2} + 7x + 2 = 0$
We compare this equation with $a{x^2} + bx + c = 0$
$a{x^2} + bx + c = 0$ is a general quadratic equation $a = 3,b = 7,c = 2$
We check the nature of the root first.
We find the discriminant for this
$D = {b^2} - 4ac$
If $D > 0$, then roots are real and distinct.
If $D = 0$, then roots are real and equal.
If $D < 0$, then roots are imaginary.
Now $D = {\left( 7 \right)^2} - 4 \times 3 \times 2$
$ = 49 - 24$
$D = 25 > 0$
$D > 0$, so roots are real and distinct.
The quadratic formula is given as
$x = \dfrac{{b \pm \sqrt D }}{{2a}}$, $D$ is discriminant
So $x = \dfrac{{ - 7 \pm \sqrt {25} }}{{2 \times 3}}$ \[ = \dfrac{{ - 7 \pm 5}}{6}\]
$x = \dfrac{{ - 7 \pm 5}}{6}$
$x = \dfrac{{ - 7 + 5}}{6},\dfrac{{ - 7 - 5}}{6}$
$x = \dfrac{{ - 2}}{6},\dfrac{{ - 12}}{6}$
$x = \dfrac{{ - 1}}{3}, - 3$
So, roots of the quadratic equation $3{x^2} + 7x + 2 = 0$ is $\dfrac{{ - 1}}{3}, - 3$
Note: Quadratic Equation : A quadratic equation is any equation that can be rearranged in standard form as there it represents unknown and a, b, c represent known number. Where $a \ne D$. If $a = 0$ then it will not be a quadratic equation.
The quadratic equation is the equation of parabola. There are two solutions of the quadratic equation; these solutions are called roots equations and these solutions are called roots. The roots may be real distinct, real and same and not real.
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