Question

How do you find the solution to the quadratic equation $3{x^2} + 24x - 9 = 0$?

Answer 1

Hint: First take $3$ common from the given equation and then divide both sides of the equation by $3$. Next, compare the given quadratic equation to the standard quadratic equation and find the value of numbers $a$, $b$ and $c$ in the given equation. Then, substitute the values of $a$, $b$ and $c$ in the formula of discriminant and find the discriminant of the given equation. Finally, put the values of $a$, $b$ and $D$ in the roots of the quadratic equation formula and get the desired result.

Formula used: Chain Rule:
The quantity $D = {b^2} - 4ac$ is known as the discriminant of the equation $a{x^2} + bx + c = 0$ and its roots are given by
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ or $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step by step solution:
We know that an equation of the form $a{x^2} + bx + c = 0$, $a,b,c,x \in R$, is called a Real Quadratic Equation.
The numbers $a$, $b$ and $c$ are called the coefficients of the equation.
The quantity $D = {b^2} - 4ac$ is known as the discriminant of the equation $a{x^2} + bx + c = 0$ and its roots are given by
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ or $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
So, first we will take $3$ common from the given equation.
$ \Rightarrow 3\left( {{x^2} + 8x - 3} \right) = 0$
Divide both sides of the equation by $3$.
$ \Rightarrow {x^2} + 8x - 3 = 0$
Next, compare ${x^2} + 8x - 3 = 0$ quadratic equation to standard quadratic equation and find the value of numbers $a$, $b$ and $c$.
Comparing ${x^2} + 8x - 3 = 0$ with $a{x^2} + bx + c = 0$, we get
$a = 1$, $b = 8$ and $c = - 3$
Now, substitute the values of $a$, $b$ and $c$ in $D = {b^2} - 4ac$ and find the discriminant of the given equation.
$ \Rightarrow D = {\left( 8 \right)^2} - 4\left( 1 \right)\left( { - 3} \right)$
After simplifying the result, we get
$ \Rightarrow D = 76$
Which means the given equation has real roots.
Now putting the values of $a$, $b$ and $D$ in $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$, we get
$ \Rightarrow x = \dfrac{{ - 8 \pm 2\sqrt {19} }}{{2 \times 1}}$
Divide numerator and denominator by $2$, we get
$ \Rightarrow x = - 4 \pm \sqrt {19} $
So, $x = - 4 + \sqrt {19} $ and $x = - 4 - \sqrt {19} $ are roots/solutions of equation ${x^2} - \dfrac{2}{3}x - \dfrac{2}{3}$.

Therefore, the solution to the quadratic equation $3{x^2} + 24x - 9 = 0$ are $x = - 4 \pm \sqrt {19} $.

Note: We can check whether $x = - 4 + \sqrt {19} $ and $x = - 4 - \sqrt {19} $are roots/solutions of equation $3{x^2} + 24x - 9 = 0$ by putting the value of $x$ in given equation.
Putting $x = - 4 + \sqrt {19} $ in LHS of equation $3{x^2} + 24x - 9 = 0$.
\[{\text{LHS}} = 3{\left( { - 4 + \sqrt {19} } \right)^2} + 24\left( { - 4 + \sqrt {19} } \right) - 9\]
On simplification, we get
\[{\text{LHS}} = 3{\left( { - 4 + \sqrt {19} } \right)^2} + 24\left( { - 4 + \sqrt {19} } \right)\]
On squaring the term and we get
\[ \Rightarrow 3\left( {16 + 19 - 8\sqrt {19} } \right) - 96 + 24\sqrt {19} - 9\]
Let us add the term and we get
\[ \Rightarrow 3\left( {35 - 8\sqrt {19} } \right) + 24\sqrt {19} - 105\]
On multiply we get,
\[ \Rightarrow 105 - 24\sqrt {19} + 24\sqrt {19} - 105\]
On simplify the term and we get,
\[ \Rightarrow 0\]
Thus, $x = - 4 + \sqrt {19} $ is a solution of equation $3{x^2} + 24x - 9 = 0$.
Putting $x = - 4 - \sqrt {19} $ in LHS of equation $3{x^2} + 24x - 9 = 0$.
\[{\text{LHS}} = 3{\left( { - 4 - \sqrt {19} } \right)^2} + 24\left( { - 4 - \sqrt {19} } \right) - 9\]
On simplification, we get
\[{\text{LHS}} = 3{\left( { - 4 - \sqrt {19} } \right)^2} + 24\left( { - 4 - \sqrt {19} } \right) - 9\]
On squaring the term and we get
\[ \Rightarrow 3\left( {16 + 19 + 8\sqrt {19} } \right) - 96 - 24\sqrt {19} - 9\]
Let us add the term and we get,
\[ \Rightarrow 3\left( {35 + 8\sqrt {19} } \right) - 24\sqrt {19} - 105\]
On multiply the term and we get,
\[ \Rightarrow 105 + 24\sqrt {19} - 24\sqrt {19} - 105\]
On simplify the term and we get
\[ \Rightarrow 0\]
\[ \Rightarrow {\text{RHS}}\]
Thus, $x = - 4 - \sqrt {19} $ is a solution of equation $3{x^2} + 24x - 9 = 0$.
Therefore, the solution to the quadratic equation $3{x^2} + 24x - 9 = 0$ are $x = - 4 \pm \sqrt {19} $.