Question

How do you solve $4{x^2} + 7 = 23$?

Answer 1

Hint: We will first of all, write the general quadratic equation and the formula for its roots and then on comparing put the values as in formula and thus we have the required roots.

Complete step by step answer:
We are given that we are required to solve $4{x^2} + 7 = 23$ using the quadratic formula.
We can write this equation as: $4{x^2} - 16 = 0$.
The general quadratic equation is given by $a{x^2} + bx + c = 0$, where a, b and c are the constants.
The roots are this equation is given by the following expression with us:-
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Comparing the general quadratic equation, we have: a = 4, b = 0 and c = - 16.
Therefore, the roots of the equation $4{x^2} - 16 = 0$ are given by:-
$ \Rightarrow x = \dfrac{{ - 0 \pm \sqrt {{{(0)}^2} - 4(4)( - 16)} }}{{2(4)}}$
Simplifying the calculations, we get the following equation with us:-
$ \Rightarrow x = \dfrac{{0 \pm 16}}{8}$
Simplifying the calculations further, we get the following equation with us:-
$ \Rightarrow x = \dfrac{{ \pm 16}}{8}$
Crossing – off 8 from both the numerator and denominator, we will get:-
The possible values of x as $ \pm 2$.

Note:
The students must note that there is an alternate way to do the same question, it is given as follows if It is not mentioned that we need to use quadratics formula.
We are given that we are required to solve $4{x^2} + 7 = 23$ .
We can write this equation as: $4{x^2} - 16 = 0$.
Since we know that we have an identity given by the following expression with us:-
$ \Rightarrow {a^2} - {b^2} = (a - b)(a + b)$
We know we can write the above equation as ${\left( {2x} \right)^2} - {(4)^2} = 0$.
We can further write the above expression ${\left( {2x} \right)^2} - {(4)^2} = 0$ as:
$ \Rightarrow \left( {2x - 4} \right)\left( {2x + 4} \right) = 0$
Thus, we have it as: If a.b = 0, then either a = 0 or b = 0. Here, we just replaced a by (2x – 4) and b by (2x + 4). So, either (2x – 4) = 0 or (2x + 4) = 0
Hence, the possible values of x as $ \pm 2$.