Question

splitting the middle top of 2 X square + X - 4=0​

Answer 1

Answer:

$x=-[\frac{1-\sqrt{33}}{4}] \ or\ x=-[\frac{1+\sqrt{33}}{4}]$

Step-by-step explanation:

We know that the equation of last term is given by

$LastTerm = \frac{(Middle Term)^{2}}{4\times(First Term)}$

 $\therefore \ LastTerm = \frac{{x}^2}{4\times(2x^2)}=\frac{1}{8}$

$\therefore2{x}^2+x-4+\frac{1}{8}-\frac{1}{8}=0\\\therefore(\sqrt{2}x)^2+2\times(\sqrt2x)\times(\frac{1}{2\sqrt2})+\frac{1}{8}-\frac{33}{8}=0\\\therefore(\sqrt2x+\frac{1}{2\sqrt2})^2-(\frac{\sqrt{33}}{2\sqrt2})^2=0\\\therefore(\sqrt2x+\frac{1}{2\sqrt2}-\frac{\sqrt{33}}{2\sqrt2})(\sqrt2x+\frac{1}{2\sqrt2}+\frac{\sqrt{33}}{2\sqrt2})=0\\\therefore(\sqrt2x+\frac{1-\sqrt{33}}{2\sqrt2})(\sqrt2x+\frac{1+\sqrt{33}}{2\sqrt2})=0$

$\\\therefore\sqrt2x+\frac{1-\sqrt{33}}{2\sqrt2}=0 \ or\ \sqrt2x+\frac{1+\sqrt{33}}{2\sqrt2}=0\\\\\therefore \ x=-[\frac{1-\sqrt{33}}{4}] \ or\ x=-[\frac{1+\sqrt{33}}{4}]$