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Maths-

General

Easy

Question

Solve $open parentheses x squared plus 4 x plus 5 close parentheses to the power of 3 over 2 end exponent plus 1 equals 0$

hint Hint:

Rearrange the equation and then solve.

The correct answer is: x = - 2

Complete step by step solution:
Here we have the equation $open parentheses x squared plus 4 x plus 5 close parentheses to the power of 3 over 2 end exponent plus 1 equals 0$
On subtracting 1 from both the sides, we get $open parentheses x squared plus 4 x plus 5 close parentheses to the power of 3 over 2 end exponent equals negative 1$
To make the power of the equation as 1, we have to raise the equation by $2 over 3$
$not stretchy rightwards double arrow open parentheses x squared plus 4 x plus 5 close parentheses to the power of 3 over 2 cross times 2 over 3 end exponent equals negative 1 to the power of 2 over 3 end exponent$
$not stretchy rightwards double arrow open parentheses x squared plus 4 x plus 5 close parentheses to the power of negative 1 end exponent equals cube root of left parenthesis negative 1 right parenthesis squared end root$
$not stretchy rightwards double arrow x squared plus 4 x plus 5 equals cube root of 1$
$not stretchy rightwards double arrow x squared plus 4 x plus 5 equals 1$
On rearranging, we get $x squared plus 4 x plus 4 equals 0$
Now, this is a quadratic equation with a = 1, b = 4, c = 4
Roots can be found with, $x equals fraction numerator negative b plus-or-minus square root of b squared minus 4 a c end root over denominator 2 a end fraction$
$x equals fraction numerator negative 4 plus-or-minus square root of 16 minus 4 cross times 1 cross times 4 end root over denominator 2 cross times 1 end fraction$
On solving, we get x = - 2

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