Question

IT IS URGENT !!! PLEASE HELP, IT'S FOR TODAY !!!

3. Solve the following quadratic equations.

4x2 = 0
4x2-3x = 0
x2 + 5x + 6 = 0
n2-5n + 4 = 0
3x2 + x-2 = 0
6x2 + 3x-3 = 0

4. Given the following rectangle, determining obligations should be the value of x that allows to obtain the area indicated.

Answer 1

Part 3)

a)
`x=0`

b)
`x=0` and
`x=(3)/(4)`

c)
`x=-3` and
`x=-2`

d)
`n=1` and
`n=4`

e)
`x=-1` and
`x=(2)/(3)`

f)
`x=-1` and
`x=(1)/(2)`

Part 4)
`x=10\ cm`

Explanation:

Part 3) Solve the following quadratic equations.

case a) we have

`4x^2=0`

Divide by 4 both sides

`x^2=0`

take square root both sides

`x=0`

case b) we have

`4x^2-3x=0`

Factor x

`x(4x-3)=0`

so

One solution is

`x=0`

Second solution is

`(4x-3)=0`

`x=(3)/(4)`

case c) we have

`x^2+5x+6=0`

The formula to solve a quadratic equation of the form

`ax^(2) +bx+c=0`

is equal to

`x=\frac{-b\pm\sqrt{b^(2)-4ac}} {2a}`

in this problem we have

`x^2+5x+6=0`

so

`a=1\\b=5\\c=6`

substitute in the formula

`x=\frac{-5\pm\sqrt{5^(2)-4(1)(6)}} {2(1)}`

`x=\frac{-5\pm√(1)} {2}`

`x=\frac{-5\pm1} {2}`

therefore

`x=\frac{-5+1} {2}=-2`

`x=\frac{-5-1} {2}=-3`

case d) we have

`n^2-5n+4=0`

The formula to solve a quadratic equation of the form

`an^(2) +bn+c=0`

is equal to

`n=\frac{-b\pm\sqrt{b^(2)-4ac}} {2a}`

in this problem we have

`n^2-5n+4=0`

so

`a=1\\b=-5\\c=4`

substitute in the formula

`n=\frac{-(-5)\pm\sqrt{-5^(2)-4(1)(4)}} {2(1)}`

`n=\frac{5\pm√(9)} {2}`

`n=\frac{5\pm3} {2}`

therefore

`n=\frac{5+3} {2}=4`

`n=\frac{5-3} {2}=1`

case e) we have

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

The formula to solve a quadratic equation of the form

`ax^(2) +bx+c=0`

is equal to

`x=\frac{-b\pm\sqrt{b^(2)-4ac}} {2a}`

in this problem we have

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

so

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

substitute in the formula

`x=\frac{-1\pm\sqrt{1^(2)-4(3)(-2)}} {2(3)}`

`x=\frac{-1\pm√(25)} {6}`

`x=\frac{-1\pm5} {6}`

therefore

`x=\frac{-1+5} {6}=(2)/(3)`

`x=\frac{-1-5} {6}=-1`

case f) we have

`6x^2+3x-3=0`

The formula to solve a quadratic equation of the form

`ax^(2) +bx+c=0`

is equal to

`x=\frac{-b\pm\sqrt{b^(2)-4ac}} {2a}`

in this problem we have

`6x^2+3x-3=0`

so

`a=6\\b=3\\c=-3`

substitute in the formula

`x=\frac{-3\pm\sqrt{3^(2)-4(6)(-3)}} {2(6)}`

`x=\frac{-3\pm√(81)} {12}`

`x=\frac{-3\pm9} {12}`

therefore

`x=(-3+9)/(12)=(1)/(2)`

`x=(-3-9)/(12)=-1`

Part 4) we know that

The area of rectangle is equal to

`A=LW`

we have

`A=66\ cm^2\\L=(x+1)\ cm\\W=(x-4)\ cm`

substitute

`66=(x+1)(x-4)`

solve for x

Apply distributive property right side

`66=x^2-4x+x-4\\x^2-3x-70=0`

solve the quadratic equation by formula

we have

`a=1\\b=-3\\c=-70`

substitute in the formula

`x=\frac{-(-3)\pm\sqrt{-3^(2)-4(1)(-70)}} {2(1)}`

`x=\frac{3\pm√(289)} {2}`

`x=\frac{3\pm17} {2}`

therefore

`x=\frac{3+17} {2}=10`

`x=\frac{3-17} {2}=-7` ----> the value of x cannot be negative

so

The solution is x=10 cm

`L=(10+1)=11\ cm\\W=10-4=6\ cm`