Respuesta :
Here all the given equations are in the form of
[tex]ax^2+bx+c=0[/tex]
The value of [tex]x[/tex] can be determined by the perfect square method as follows:
[tex]ax^2+bx+c=0[/tex]
[tex]\Rightarrow x^2+\frac{b}{a}+\frac{c}{a}=0[/tex]
[tex]\Rightarrow x^2+2\times\frac{b}{2a}+\left(\frac{b}{2a}\right)^2-\left(\frac{b}{2a}\right)^2+\frac{c}{a}=0[/tex]
[tex]\Rightarrow \left(x+\frac{b}{2a}\right)^2=\left(\frac{b}{2a}\right)^2-\frac{c}{a}[/tex]
[tex]\Rightarrow x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\;\cdots(i)[/tex]
Now, for [tex]x^2+3x-4=0[/tex]
[tex]a=1, b=3, c=-4[/tex]
So, from equation (i)
[tex]x=\frac{-3\pm\sqrt{3^2-4\times1\times(-4)}}{2\times1}[/tex]
[tex]\Rightarrow x=\frac{3\pm\sqrt{9+16}}{2}[/tex]
[tex]\Rightarrow x=\frac{3\pm5}{2}[/tex]
[tex]\Rightarrow x=\frac{3+5}{2},\frac{3-5}{2}[/tex]
[tex]\Rightarrow x= 4, -1[/tex]
Similarly, for [tex]9x^2+12x+4=0[/tex]
[tex]a=9, b=12, c=4[/tex]
So, from equation (i)
[tex]x=\frac{-12\pm\sqrt{12^2-4\times9\times4}}{2\times9}[/tex]
[tex]\Rightarrow x=\frac{-12\pm0}{18}[/tex]
[tex]\Rightarrow x=\frac{-2}{3}[/tex]
For [tex]x^2-2x-3=0[/tex]
[tex]a=1, b=-2, c=-3[/tex]
[tex]x=\frac{-(-2)\pm\sqrt{(-2)^2-4\times1\times(-3)}}{2\times1}[/tex]
[tex]\Rightarrow x=\frac{2\pm\sqrt{16}}{2}[/tex]
[tex]\Rightarrow x=\frac{2\pm4}{2}[/tex]
[tex]\Rightarrow x=\frac{6}{2},\frac{-2}{2}[/tex]
[tex]\Rightarrow x=3.00, -1.00[/tex]
For [tex]x^2-6x+9=0[/tex]
[tex]a=1, b=-6, c=9[/tex]
[tex]x=\frac{-(-6)\pm\sqrt{(-6)^2-4\times1\times9}}{2\times1}[/tex]
[tex]\Rightarrow x=\frac{6\pm0}{2}[/tex]
[tex]\Rightarrow x=3.00[/tex]
For [tex]2x^2-4x-3=0[/tex]
[tex]a=2, b=-4, c=-3[/tex]
[tex]x=\frac{-(-4)\pm\sqrt{(-4)^2-4\times2\times(-3)}}{2\times2}[/tex]
[tex]\Rightarrow x=\frac{4\pm\sqrt{40}}{4}[/tex]
[tex]\Rightarrow x=\frac{4\pm2\sqrt{10}}{4}[/tex]
[tex]\Rightarrow x=\frac{4+2\sqrt{10}}{4},\frac{4-2\sqrt{10}}{4}[/tex]
[tex]\Rightarrow x=2.58, -0.58[/tex]
