Question

How do you solve $9{x^2} - 12x + 4 = - 3$ ?

Answer 1

Hint:The given equation is a quadratic equation in one variable $x$. It should first be changed into the general form of a quadratic equation, which is given as $a{x^2} + bx + c = 0$. We will then use the coefficients to find the value of discriminant $D = {b^2} - 4ac$. If $D \geqslant 0$, we use the quadratic formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$to solve for $x$.

Complete step by step solution:
We have to solve the given equation $9{x^2} - 12x + 4 = - 3$ using the quadratic formula$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.
To find the value of $x$, we have to put the values of $a$, $b$ and $c$ in the quadratic formula. To get the values of $a$, $b$ and $c$ from the given equation, we rearrange the equation and compare it with the general form of the quadratic equation.
$
9{x^2} - 12x + 4 = - 3 \\
\Rightarrow 9{x^2} - 12x + 4 + 3 = 0 \\
\Rightarrow 9{x^2} - 12x + 7 = 0 \\
$
General form of quadratic equation is written in the form of $a{x^2} + bx + c = 0$, where $a$ is the coefficient of ${x^2}$, $b$ is the coefficient of $x$ and $c$ is the constant term. The RHS is $0$. On comparing the above rearranged equation with the general form, we observe that:
Coefficient $a$ of ${x^2}$ is $9$,
Coefficient $b$ of $x$ is $ - 12$,
and the constant term $c$ is $7$.
Thus, $a = 9$, $b = - 12$ and $c = 7$.
Now we check the value of discriminant $D$ by using the above values. $D$ is given by,
$D = {b^2} - 4ac$
If $D > 0$, the two roots will be real and distinct.
If $D = 0$, the two roots will be real and equal.
If $D < 0$, the two roots will be imaginary and distinct.
For $a = 9$, $b = - 12$and $c = 7$, we have:
$
D = {b^2} - 4ac \\
\Rightarrow D = {( - 12)^2} - 4 \times 9 \times 7 \\
\Rightarrow D = 144 - 252 \\
\Rightarrow D = - 108 \\
$
Since, $D$ is negative here, the roots will be imaginary in nature. Therefore, there is no real solution for the given equation.

Note: When solving a quadratic equation, discriminant of the equation should always be calculated to check if the roots are real or not. In case, $D < 0$, the real solution of the equation would not be possible. We can see that if $D < 0$, we get a negative term under the roots in the quadratic formula, the real solution of which is not possible.