Answer:
Please check the explanation.
Step-by-step explanation:
Given the equation
-2x² = 4-3 (x + 1)
-2x² = 4-3x-3
-2x² = -3x -7
0 = 2x² -3x -7
We know that the degree of the equation is the highest power of x variable in the given equation.
In the equation 0 = 2x² -3x -7 the highest power of x variable in the given equation is 2.
Thus, the degree of the equation is 2.
Also in the equation 0 = 2x² -3x -7, the unknown variable is 'x'.
Let us determine the value 'x'
2x² -3x -7 = 0
Add 7 to both sides
[tex]2x^2-3x-7+7=0+7[/tex]
[tex]2x^2-3x=7[/tex]
Divide both sides by 2
[tex]\frac{2x^2-3x}{2}=\frac{7}{2}[/tex]
[tex]x^2-\frac{3x}{2}=\frac{7}{2}[/tex]
Add (-3/4)² to both sides
[tex]x^2-\frac{3x}{2}+\left(-\frac{3}{4}\right)^2=\frac{7}{2}+\left(-\frac{3}{4}\right)^2[/tex]
[tex]x^2-\frac{3x}{2}+\left(-\frac{3}{4}\right)^2=\frac{65}{16}[/tex]
[tex]\left(x-\frac{3}{4}\right)^2=\frac{65}{16}[/tex]
[tex]\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}[/tex]
solving
[tex]x-\frac{3}{4}=\sqrt{\frac{65}{16}}[/tex]
[tex]x-\frac{3}{4}=\frac{\sqrt{65}}{\sqrt{16}}[/tex]
[tex]x-\frac{3}{4}=\frac{\sqrt{65}}{4}[/tex]
Add 3/4 to both sides
[tex]x-\frac{3}{4}+\frac{3}{4}=\frac{\sqrt{65}}{4}+\frac{3}{4}[/tex]
[tex]x=\frac{\sqrt{65}+3}{4}[/tex]
similarly solving
[tex]x-\frac{3}{4}=-\sqrt{\frac{65}{16}}[/tex]
[tex]x=\frac{-\sqrt{65}+3}{4}[/tex]
So the solution of the equation will have the values of x such as:
[tex]x=\frac{\sqrt{65}+3}{4},\:x=\frac{-\sqrt{65}+3}{4}[/tex]
