Question

How do you solve ${\left( {3x - 2} \right)^2} - 7 = 0$?

Answer 1

Hint: First, open the bracket using the algebraic identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and simplify. Next, compare the 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: 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 $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
First, open the bracket using the algebraic identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and simplify.
$ \Rightarrow 9{x^2} - 12x + 4 - 7 = 0$
$ \Rightarrow 9{x^2} - 12x - 3 = 0$
Now, compare $9{x^2} - 12x - 3 = 0$ quadratic equation to standard quadratic equation and find the value of numbers $a$, $b$ and $c$.
Comparing $9{x^2} - 12x - 3 = 0$ with $a{x^2} + bx + c = 0$, we get
$a = 9$, $b = - 12$ 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.
$D = {\left( { - 12} \right)^2} - 4\left( 9 \right)\left( { - 3} \right)$
After simplifying the result, we get
$ \Rightarrow D = 144 + 108$
$ \Rightarrow D = 252$
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
$x = \dfrac{{ - \left( { - 12} \right) \pm 6\sqrt 7 }}{{2 \times 9}}$
Divide numerator and denominator by $6$, we get
\[x = \dfrac{2}{3} \pm \dfrac{{\sqrt 7 }}{3}\]
$ \Rightarrow x = \dfrac{2}{3} + \dfrac{{\sqrt 7 }}{3}$ and $x = \dfrac{2}{3} - \dfrac{{\sqrt 7 }}{3}$
So, $x = \dfrac{2}{3} + \dfrac{{\sqrt 7 }}{3}$ and $x = \dfrac{2}{3} - \dfrac{{\sqrt 7 }}{3}$ are roots/solutions of equation ${\left( {3x - 2} \right)^2} - 7 = 0$.

Therefore, the solutions to the quadratic equation ${\left( {3x - 2} \right)^2} - 7 = 0$ are $x = \dfrac{2}{3} + \dfrac{{\sqrt 7 }}{3}$ and $x = \dfrac{2}{3} - \dfrac{{\sqrt 7 }}{3}$.

Note: We can also find the solution of the quadratic equation ${\left( {3x - 2} \right)^2} = 7$ by taking the square root of each side of the equation.
$3x - 2 = \pm \sqrt 7 $
$ \Rightarrow 3x - 2 = \sqrt 7 $ and $3x - 2 = - \sqrt 7 $
Therefore, $x = \dfrac{{2 + \sqrt 7 }}{3}$ and $x = \dfrac{{2 - \sqrt 7 }}{3}$.
Therefore, the solutions to the quadratic equation ${\left( {3x - 2} \right)^2} - 7 = 0$ are $x = \dfrac{2}{3} + \dfrac{{\sqrt 7 }}{3}$ and $x = \dfrac{2}{3} - \dfrac{{\sqrt 7 }}{3}$.