# Solving a quadratic equation using the “splitting the middle term” method.

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

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

• While I think I know what you mean, it might help you to clarify what you mean by this "splitting the middle term" method. – Eevee Trainer Jan 21 at 18:08
• Factorise the above polynomial and find its roots. – Mursaleen Salroo Jan 21 at 18:09
• There are multiple methods to factorize polynomials - you'll have to clarify what this specific method entails if you want us to be best able to help you. – Eevee Trainer Jan 21 at 18:10
• The polynomial x2 + cx + d, where a + b = c and ab = d, can be factorized into (x + a)(x + b). – Mursaleen Salroo Jan 21 at 18:13
• Your polynomial cannot be factored with respect to the rational numbers. Had you instead used $5x^2 - 12x + 4$, you could split the linear term as $5x^2 - 10x - 2x + 4$ (find two numbers with product $5 \cdot 4 = 20$ with sum $-12$) to obtain the factorization \begin{align*} 5x^2 - 12x + 4 & = 5x^2 - 10x - 2x + 4 && \text{split the linear term}\\ & = 5x(x - 2) - 2(x - 2) && \text{factor by grouping} \\ & = (5x - 2)(x - 2) && \text{extract the common factor} \end{align*} – N. F. Taussig Jan 21 at 18:18

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