Question

Solve showing steps. Use Insert equation to get the symbol and Ctrl with . For exponents x2 + 3x - 4 = 0 9x2 + 12x + 4 =0 (leave answer as a fraction) x2 - 2x - 3 = 0 x2 - 6x + 9 = 0 2x2 - 4x - 3 = 0 (round to two decimal places)*

Answer 1

Here all the given equations are in the form of

$ax^2+bx+c=0$

The value of $x$ can be determined by the perfect square method as follows:

$ax^2+bx+c=0$

$\Rightarrow x^2+\frac{b}{a}+\frac{c}{a}=0$

$\Rightarrow x^2+2\times\frac{b}{2a}+\left(\frac{b}{2a}\right)^2-\left(\frac{b}{2a}\right)^2+\frac{c}{a}=0$

$\Rightarrow \left(x+\frac{b}{2a}\right)^2=\left(\frac{b}{2a}\right)^2-\frac{c}{a}$

$\Rightarrow x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\;\cdots(i)$

Now, for $x^2+3x-4=0$

$a=1, b=3, c=-4$

So, from equation (i)

$x=\frac{-3\pm\sqrt{3^2-4\times1\times(-4)}}{2\times1}$

$\Rightarrow x=\frac{3\pm\sqrt{9+16}}{2}$

$\Rightarrow x=\frac{3\pm5}{2}$

$\Rightarrow x=\frac{3+5}{2},\frac{3-5}{2}$

$\Rightarrow x= 4, -1$

Similarly, for $9x^2+12x+4=0$

$a=9, b=12, c=4$

So, from equation (i)

$x=\frac{-12\pm\sqrt{12^2-4\times9\times4}}{2\times9}$

$\Rightarrow x=\frac{-12\pm0}{18}$

$\Rightarrow x=\frac{-2}{3}$

For $x^2-2x-3=0$

$a=1, b=-2, c=-3$

$x=\frac{-(-2)\pm\sqrt{(-2)^2-4\times1\times(-3)}}{2\times1}$

$\Rightarrow x=\frac{2\pm\sqrt{16}}{2}$

$\Rightarrow x=\frac{2\pm4}{2}$

$\Rightarrow x=\frac{6}{2},\frac{-2}{2}$

$\Rightarrow x=3.00, -1.00$

For $x^2-6x+9=0$

$a=1, b=-6, c=9$

$x=\frac{-(-6)\pm\sqrt{(-6)^2-4\times1\times9}}{2\times1}$

$\Rightarrow x=\frac{6\pm0}{2}$

$\Rightarrow x=3.00$

For $2x^2-4x-3=0$

$a=2, b=-4, c=-3$

$x=\frac{-(-4)\pm\sqrt{(-4)^2-4\times2\times(-3)}}{2\times2}$

$\Rightarrow x=\frac{4\pm\sqrt{40}}{4}$

$\Rightarrow x=\frac{4\pm2\sqrt{10}}{4}$

$\Rightarrow x=\frac{4+2\sqrt{10}}{4},\frac{4-2\sqrt{10}}{4}$

$\Rightarrow x=2.58, -0.58$