A precalculus question help mee?

Question

Using quadratci formula find roots in somplest radical form
x^2=6x+1
2x^2-9=0
x^2=6x+31
x(x+4)=1

USing quadratics, x must = simplest radical form
x^2=6x+11
3x^2-5=0
x(x+8)=34

How would i do that pleas explain step by step

Answer 1

for ax^2+bx+c=0 x is given by : x= minus b+or- the SQRT of b squared -4*ac with everything so far divided by 2a.
The first one : X^2-6x-1=0, a=1,b=-6,c=-1 just substitute it in . One of the factors inside the SQRT sign might be a square e.g.SQRT12=Sqrt(4*3)=2sqrt(3)= simplest radical form.

Answer 2

Assuming you didn't mistype anything

x^2 = 6x + 1
x^2 - 6x - 1 = 0

x = (6 ± sqrt(36 - 4(1)(-1)))/(2(1))
x = (6 ± sqrt(36 + 4))/2
x = (6 ± sqrt(40))/2
x = (6 ± 2sqrt(10))/2

x = 3 ± sqrt(10)

---------------------------------------

2x^2 - 9 = 0

x = (±sqrt(-4(2)(-9)))/(2(2))
x = (±sqrt(36 * 2))/4
x = (±6sqrt(2))/4
x = ±(3/2)sqrt(2)

---------------------------------------

x^2 = 6x + 31
x^2 - 6x - 31 = 0

x = (6 ± sqrt(36 - 4(1)(-31)))/(2(1))
x = (6 ± sqrt(36 + 124))/2
x = (6 ± sqrt(160))/2
x = (6 ± 4sqrt(10))/2
x = 3 ± 2sqrt(10)

-----------------------------------------

x(x + 4) = 1
x^2 + 4x = 1
x^2 + 4x - 1 = 0

x = (-4 ± sqrt(16 - 4(1)(-1)))/(2(1))
x = (-4 ± sqrt(16 + 4))/2
x = (-4 ± sqrt(20))/2
x = (-4 ± 4sqrt(5))/2
x = -2 ± 2sqrt(5)

------------------------------------

x^2 = 6x + 11
x^2 - 6x - 11 = 0

x = (6 ± sqrt(36 - 4(1)(-11)))/(2(1))
x = (6 ± sqrt(36 + 44))/2
x = (6 ± sqrt(80))/2
x = (6 ± 4sqrt(5))/2
x = 3 ± 2sqrt(5)

------------------------------------

3x^2 - 5 = 0

x = (±sqrt(-4(3)(-5)))/(2(3))
x = (±sqrt(4 * 15))/6
x = (±2sqrt(15))/6
x = ±(1/3)sqrt(15)

--------------------------------------

x(x + 8) = 34
x^2 + 8x = 34
x^2 + 8x - 34 = 0

x = (-8 ± sqrt(64 - 4(1)(-34)))/(2(1))
x = (-8 ± sqrt(64 + 136))/2
x = (-8 ± sqrt(200))/2
x = (-8 ± 10sqrt(2))/2
x = -4 ± 5sqrt(2)

Answer 3

quadratice formula is the opposite of b plus or minus the square root of b^2 minus 4ac all over 2a.
so you put the equation equal to 0:
0=x^2-6x-1 where 0=ax^2+bx+c
then stick it the formula:
6{plus or minus} {square root of}((-6)^2-4(1)(-1))/2(1)
simplify:
6{plus or minus} {square root of}(36+4)/2
6{plus or minus} {square root of}(40)/2
that would make your equations (6+{root of}40)/2 and (6- {root of} 40)/2
there is a perfect square of 4 in 40 so your radicals simplified are 2{10}
so your equations can be simplified to 3+{root of}10 and 3- {root of}10