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Question: How do you solve \({x^2} + 5 = 41\)?...

How do you solve x2+5=41{x^2} + 5 = 41?

Explanation

Solution

First of all we will take the constant term on the right and then do the calculation as required. After that we will just find the required answer.

Complete step-by-step answer:
We are given that we are required to solve x2+5=41{x^2} + 5 = 41.
On taking the 5 from addition in the left hand side to subtraction in the right hand side, we will then obtain the following expression:-
⇒x2=41−5\Rightarrow {x^2} = 41 - 5
As we can see in the above expression, we are required to simplify the calculations and get the following expression:-
⇒x2=36\Rightarrow {x^2} = 36
Since, we know that square of both 6 and -6 is 36.

Hence, the required answer is x = -6, 6.

Note:
The students must notice that the given equation is a quadratic equation. So, it will always have two roots and the nature of the roots depend upon the discriminant of the equation. If the discriminant of an equation is greater than 0, it has real and distinct roots, if the discriminant equals 0, then it has real and equal roots and if the discriminant is less than 0, then it has imaginary roots which exist in conjugate pairs.
Reading the information above and seeing the roots, we can say that the discriminant of the above equation must be greater than 0.
Let us look at the alternate way that is by using the quadratics formula to solve the same question:-
We have x2+5=41{x^2} + 5 = 41.
We can write the given equation as x2−36=0{x^2} - 36 = 0
Now, if we compare it to the general quadratic equation, which is given by ax2+bx+c=0a{x^2} + bx + c = 0 whose roots are given by:- x=−b±b2−4ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}, we get a = 1, b = 0 and c = -36.
Let us put these values in the formula mentioned above so that we get the following roots:
⇒x=−0±02−4×1×(−36)2×1\Rightarrow x = \dfrac{{ - 0 \pm \sqrt {{0^2} - 4 \times 1 \times \left( { - 36} \right)} }}{{2 \times 1}}
Simplifying the calculations a bit to get:-
⇒x=±1442\Rightarrow x = \dfrac{{ \pm \sqrt {144} }}{2}
We know that square of 12 is 144.
So, we get the roots as x=±122x = \dfrac{{ \pm 12}}{2} that is – 6 and 6.