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+(b)/(a)+(c)/(a)=0`

`\Rightarrow x^2+2*(b)/(2a)+\left((b)/(2a)\right)^2-\left((b)/(2a)\right)^2+(c)/(a)=0`

`\Rightarrow \left(x+(b)/(2a)\right)^2=\left((b)/(2a)\right)^2-(c)/(a)`

`\Rightarrow x=(-b\pm√(b^2-4ac))/(2a)\;\cdots(i)`

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

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

So, from equation (i)

`x=(-3\pm√(3^2-4*1*(-4)))/(2*1)`

`\Rightarrow x=(3\pm√(9+16))/(2)`

`\Rightarrow x=(3\pm5)/(2)`

`\Rightarrow x=(3+5)/(2),(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=(-12\pm√(12^2-4*9*4))/(2*9)`

`\Rightarrow x=(-12\pm0)/(18)`

`\Rightarrow x=(-2)/(3)`

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

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

`x=(-(-2)\pm√((-2)^2-4*1*(-3)))/(2*1)`

`\Rightarrow x=(2\pm√(16))/(2)`

`\Rightarrow x=(2\pm4)/(2)`

`\Rightarrow x=(6)/(2),(-2)/(2)`

`\Rightarrow x=3.00, -1.00`

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

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

`x=(-(-6)\pm√((-6)^2-4*1*9))/(2*1)`

`\Rightarrow x=(6\pm0)/(2)`

`\Rightarrow x=3.00`

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

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

`x=(-(-4)\pm√((-4)^2-4*2*(-3)))/(2*2)`

`\Rightarrow x=(4\pm√(40))/(4)`

`\Rightarrow x=(4\pm2√(10))/(4)`

`\Rightarrow x=(4+2√(10))/(4),(4-2√(10))/(4)`

`\Rightarrow x=2.58, -0.58`