Question

How do you solve $21{a^2} + 5a = 4$?

Answer 1

Hint: Here we need to find the solution to the given polynomial $21{a^2} + 5a = 4$. Note that the degree of the polynomial is 2, so it is a quadratic polynomial. We convert this polynomial to the form $x{a^2} + ya + z$. So we rewrite the middle term as a sum of terms whose product is $x \cdot z$ and whose sum is y. Therefore, substitute the values of x, y, z and solve the given problem by splitting the middle term.

Complete step-by-step answer:
Given an equation of the form $21{a^2} + 5a = 4$
It is mentioned that we need to find the solution to the given polynomial.
We find the solution by factoring method.
We rewrite the given equation as,
$21{a^2} + 5a - 4 = 0$
Note that the given equation is a quadratic equation, so we factor it by splitting the middle term.
Consider an equation of the form $x{a^2} + ya + z$, where x, y, z are any real numbers.
We rewrite the middle term as a sum of two terms in such a way that their product is $x \cdot z$ and their sum is y.
In the given equation we have $x = 21$, $y = 5$ and $z = - 4$
We split the middle term $5a$ as, $5a = 12a - 7a$.
Note that their product is,
 $x \cdot z = 21 \times ( - 4)$
$ \Rightarrow x \cdot z = - 84$
$ \Rightarrow - 84 = 21 \times ( - 4)$.
Note that their sum is,
$y = 5$
$ \Rightarrow 5 = 12 - 7$.
Hence the equation $21{a^2} + 5a - 4 = 0$ can be written as,
$ \Rightarrow 21{a^2} + 12a - 7a - 4 = 0$
Factor out the greatest common factor from each group, we get,
$ \Rightarrow 3a(7a + 4) - 1(7a + 4) = 0$
Now factor the polynomial by factoring out the greatest common factor $7a + 4$, we get,
$ \Rightarrow (7a + 4)(3a - 1) = 0$
i.e. either $7a + 4 = 0$ or $3a - 1 = 0$
If $7a + 4 = 0$, then we have,
$ \Rightarrow 7a = - 4$
$ \Rightarrow a = \dfrac{{ - 4}}{7}$
If $3a - 1 = 0$, then we have,
$ \Rightarrow 3a = 1$
$ \Rightarrow a = \dfrac{1}{3}$
Hence the solution of the equation $21{a^2} + 5a = 4$ is $a = \dfrac{{ - 4}}{7}$ or $a = \dfrac{1}{3}$

Note:
Alternative method:
Given a quadratic equation $21{a^2} + 5a = 4$.
We rewrite the given equation as, $21{a^2} + 5a - 4 = 0$
Here we find the roots of the given equation and then from the roots we try to find out the factors.
This equation is of the form of $x{a^2} + ya + z$. We find the roots using the formula,
$a = \dfrac{{ - y \pm \sqrt {{y^2} - 4xz} }}{{2x}}$.
Here $x = 21$, $y = 5$ and $z = - 4$
Substituting these values in the formula we get,
$ \Rightarrow a = \dfrac{{ - 5 \pm \sqrt {{{(5)}^2} - 4 \times 21 \times ( - 4)} }}{{2 \times 21}}$
$ \Rightarrow a = \dfrac{{ - 5 \pm \sqrt {25 - 84 \times ( - 4)} }}{{42}}$
$ \Rightarrow a = \dfrac{{ - 5 \pm \sqrt {25 + 336} }}{{42}}$
$ \Rightarrow a = \dfrac{{ - 5 \pm \sqrt {361} }}{{42}}$
We know that $\sqrt {361} = 19$.
$ \Rightarrow a = \dfrac{{ - 5 \pm 19}}{{42}}$
Hence we get two roots given by,
${a_1} = \dfrac{{ - 5 + 19}}{{42}}$ and ${a_2} = \dfrac{{ - 5 - 19}}{{42}}$
$ \Rightarrow {a_1} = \dfrac{{14}}{{42}}$ and ${a_2} = - \dfrac{{24}}{{42}}$
$ \Rightarrow {a_1} = \dfrac{1}{3}$ and ${a_2} = - \dfrac{4}{7}$
Hence the solution of the equation $21{a^2} + 5a = 4$ is ${a_1} = \dfrac{1}{3}$ and ${a_2} = \dfrac{{ - 4}}{7}$.