Question

How do you solve ${\left( {x - 1} \right)^2} = 4$?

Answer 1

Hint: First, open the bracket using the algebraic identity. Then, move $4$ to the left side of the equation by subtracting $4$ from both sides of the equation. 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 answer:
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 ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$.
$ \Rightarrow {x^2} - 2x + 1 = 4$
Now, move $4$ to the left side of the equation by subtracting $4$ from both sides of the equation.
$ \Rightarrow {x^2} - 2x - 3 = 0$
Now, compare ${x^2} - 2x - 3 = 0$ quadratic equation to standard quadratic equation and find the value of numbers $a$, $b$ and $c$.
Comparing ${x^2} - 2x - 3 = 0$ with $a{x^2} + bx + c = 0$, we get
$a = 1$, $b = - 2$ 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.
$D = {\left( { - 2} \right)^2} - 4\left( 1 \right)\left( { - 3} \right)$
After simplifying the result, we get
$ \Rightarrow D = 4 + 12$
$ \Rightarrow D = 16$
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
$x = \dfrac{{ - \left( { - 2} \right) \pm 4}}{{2 \times 1}}$
Divide numerator and denominator by $2$, we get
$x = 1 \pm 2$
$ \Rightarrow x = 3$ and $x = - 1$
So, $x = - 3$ and $x = - 1$ are roots/solutions of equation ${\left( {x - 1} \right)^2} = 4$.

Therefore, the solutions to the quadratic equation ${\left( {x - 1} \right)^2} = 4$ are $x = - 3$ and $x = - 1$.

Note: We can also find the solution of the quadratic equation ${\left( {x - 1} \right)^2} = 4$ by taking the square root of each side of the equation.
$x - 1 = \pm \sqrt 4 $
Simplify the right side of the equation.
Since, $\sqrt 4 $ can be written as $2$.
Thus, $x - 1 = \pm 2$
$ \Rightarrow x - 1 = 2$ and $x - 1 = - 2$
Therefore, $x = 3$ and $x = - 1$.
Final solution: Hence, the solution to the quadratic equation ${\left( {x - 1} \right)^2} = 4$ are $x = - 3$ and $x = - 1$.