Question

How to solve ${(x - 3)^2} = - 10$?

Answer 1

we can solve the equation in two ways.
For the first, we will solve by changing the sides of constant and variable terms.
Next, we will solve Sreedhar Acharya’s formula.
In both the cases, we will get the same solution.

Complete step by step solution:
It is given that, the equation is ${(x - 3)^2} = - 10$
We have to find the solution of the given equation.
We have, ${(x - 3)^2} = - 10$
Simplifying we get,
$\Rightarrow (x - 3) = \pm \sqrt { - 10}$
We know that, $\sqrt { - 1} = i$is the imaginary number.
Simplifying again we get,
$\Rightarrow (x - 3) = \pm 10i$
Simplifying again we get,
$\Rightarrow x = 3 \pm 10i$

Hence, the solution of the equation is, $x = 3 \pm 10i$

Note: We can solve the equation in another way.
The equation can be written as,
${x^2} - 6x + 9 = - 10$
Simplifying we get,
${x^2} - 6x + 19 = 0$
By Sreedhar Acharya’s formula we know that, for a quadratic equation $a{x^2} + bx + c = 0$, the roots are, $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Comparing the given equation with the general form of quadratic formula we get,
$a = 1,b = - 6,c = 19$
So, the solution is, $\dfrac{{6 \pm \sqrt { - {6^2} - 4 \times 1 \times 19} }}{{2 \times 1}}$
Simplifying we get,
$\Rightarrow \dfrac{{6 \pm \sqrt { - 40} }}{{2 \times 1}}$
Simplifying again we get,
$\Rightarrow \dfrac{{6 \pm 2 \times 10i}}{{2 \times 1}}$
Simplifying again we get,
$3 \pm 10i$, $\sqrt { - 1} = i$is the imaginary number.
In algebra, a quadratic equation (from the Latin quadratus for "square") is any equation that can be rearranged in standard form as $a{x^2} + bx + c = 0$
Where x represents an unknown, and a, b, and c represent known numbers, where a ≠ 0. If a = 0, then the equation is linear, not quadratic, as there is no $a{x^2}$ term. The numbers a, b, and c are the coefficients of the equation and may be distinguished by calling them, respectively, the quadratic coefficient, the linear coefficient and the constant or free term.
The values of x that satisfy the equation are called solutions of the equation, and roots or zeros of the expression on its left-hand side. A quadratic equation has at most two solutions. If there is no real solution, there are two complex solutions. If there is only one solution, one says that it is a double root. A quadratic equation always has two roots, if complex roots are included and a double root is counted for two. A quadratic equation can be factored into an equivalent equation $a{x^2} + bx + c = a(x - r)(x - s)$
where r and s are the solutions for x.