Solving a quadratic equation using the "splitting the middle term" method.

Question

Use splitting the middle term method to solve the below equation. Is there a limitation to this method?

$$5b^2-16b+4=0$$

Answer 1

Your polynomial is not readily factorizable: $$D = 16(16-5) = 4^211$$ $$ x_{1,2} = \frac{8 \pm 2\sqrt{11}}{5}$$ $$5(x-8+2\sqrt{11})(x-8-2\sqrt{11})$$

Answer 2

A quadratic polynomial $ax^2 + bx + c$ with integer coefficients can be factored by splitting the linear term and grouping if the quadratic equation $ax^2 + bx + c = 0$ has rational roots.

Consider the quadratic equation $$12x^2 + 7x - 10 = 0$$ To split the linear term of $12x^2 + 7x - 10$, we must find two integers with product $12 \cdot (-10) = -120$ and sum $7$. They are $15$ and $-7$. Hence, \begin{align*} 12x^2 + 7x - 10 & = 0\\ 12x^2 + 15x - 8x - 10 & = 0 && \text{split the linear term}\\ 3x(4x + 5) - 2(4x + 5) & = 0 && \text{factor by grouping}\\ (3x - 2)(4x + 5) & = 0 && \text{extract the common factor} \end{align*} Setting each factor equal to zero yields \begin{align*} 3x - 2 & = 0 & 4x + 5 & = 0\\ 3x & = 2 & 4x & = -5\\ x & = \frac{2}{3} & x & = -\frac{5}{4} \end{align*} Notice that the roots are rational numbers since we were able to factor $12x^2 + 7x - 10$ into two factors with integer coefficients.

We could also solve the equation by using the Quadratic Formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Substituting $12$ for $a$, $7$ for $b$, and $-10$ for $c$ yields \begin{align*} x & = \frac{-7 \pm \sqrt{7^2 - 4(12)(-10)}}{2(12)}\\ & = \frac{-7 \pm \sqrt{49 + 480}}{24}\\ & = \frac{-7 \pm \sqrt{529}}{24}\\ & = \frac{-7 \pm 23}{24} \end{align*} Thus, \begin{align*} x & = \frac{-7 + 23}{24} & x & = \frac{-7 - 23}{24}\\ & = \frac{16}{24} & & = -\frac{30}{24}\\ & = \frac{2}{3} & & = -\frac{5}{4} \end{align*} The roots of the quadratic equation $5x^2 - 16x + 4 = 0$ are irrational. \begin{align*} x & = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(5)(4)}}{2(5)}\\ & = \frac{16 \pm \sqrt{256 - 80}}{10}\\ & = \frac{16 \pm \sqrt{176}}{10}\\ & = \frac{16 \pm 4\sqrt{11}}{10}\\ & = \frac{8 \pm 2\sqrt{11}}{5} \end{align*} Therefore, we cannot split its linear term into terms with integer coefficients.

If you examine these two problems, you will notice that the reason the first equation produced rational roots and the second did not is that the discriminant $\Delta = b^2 - 4ac$ of the quadratic equation $12x^2 + 7x - 10 = 0$ is a perfect square while the discriminant of the quadratic equation $5x^2 - 16x + 4$ is not.

If the discriminant of a quadratic equation $ax^2 + bx + c = 0$ with rational coefficients is a perfect square, then $\sqrt{\Delta}$ is a rational number, so the roots \begin{align*} r_1 & = \frac{-b + \sqrt{\Delta}}{2a}\\ r_2 & = \frac{-b - \sqrt{\Delta}}{2a} \end{align*} are rational numbers since the rational numbers are closed under the operations of addition, subtraction, multiplication, and division by a nonzero number (if $a = 0$, then $ax^2 + bx + c = bx + c$ is not quadratic). On the other hand, if $\Delta$ is not a perfect square, then $\sqrt{\Delta}$ is irrational, so the roots of the quadratic equation are irrational since the sum of a rational number and an irrational number is irrational, as is the quotient of an irrational number and a nonzero rational number.