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Help with algebra 2 hw?

0 votes

find the roots of 

4x^2+1=0

2x^2-5x+14=0

8x^3-27=0

asked Mar 20, 2013 in ALGEBRA 2 by linda Scholar

3 Answers

0 votes

4x2 + 1 = 0

Subtract 1 to each side

4x2 + 1 - 1 = 0 - 1

4x2 + 0 = -1

4x2 = -1

Divide each side by 4

x2 = -1 / 4

But i2 = -1

Take square root  to each side

i = (-1)

Substitute i2 = -1 in the equation x2 = -1 / 4

x2 = i2 / 4

x2 = (i / 2)2

Take square root to each side

(x)2 = [(i / 2)2]

There fore x = i / 2 or x = -i / 2

The roots of the  equation : -i / 2 , i / 2.

 

answered Mar 20, 2013 by diane Scholar
0 votes

2x2 - 5x +14 = 0

ax2 + bx + c =0 then the roots of the equation :  x = [-b + √( b2 - 4ac )] / 2a or

x =  [-b - √( b2 - 4ac )] / 2a

Substitute a = 2 , b = -5 and c = 14 in the roots of the equation x

There fore x1 = [-(-5) + √( (-5)2 - 4(2)(14) )] / 2 (2) or x2 =[ -(-5) - √( (-5)2 - 4(2)(14) )] / 2 (2)

x 1= [5 + √( 25 -8(14) )] / 4 or x2 = [5 - √( 25 -8(14) )] / 4

x1 = (5 + √(25 - 112) ) / 4 or x2 = (5 - √( 25 - 112) ) / 4

x1 = 5 + √(-87) or x2 = 5 - √(-87)

But i2 = -1

Substitute i2 = -1 in the value of x1 , x2

x1 = 5 + √(i2*87) 0r x2 = 5 - √(i2*87)

sqrt(a*b) = sqrt(a) * sqrt(b) where a,b are real numbers

x1 = 5 + √(i2)* √(87) or x1 = 5 - √(i2)* √(87)

But √(i2) = i

Substitute √(i2) = i in the roots of the equation x1 , x2

x1 = 5 + i*√(87) or x2 = 5 - i*√(87)

x1 = 5 + i√(87) or x2 = 5 - i√(87)

The roots of the equation : x1 = 5 + i√(87) , x2 = 5 - i√(87).
 

answered Mar 20, 2013 by diane Scholar

Roots of  the equation 2x2 - 5x +14 = 0 is [5 + i√(87)]/4 or  [5 - i√(87)]/4.

0 votes

8x3 - 27 = 0

Add 27 to each side

8x3 -27 + 27 = 0 + 27

8x3 - 0 = 27

8x3 = 27

Divide each side by 8

8x3 / 8 = 27 / 8

x3 = 27 / 8

There fore x3 = (3 / 2)3

Take cube root  to each side

(x3)1/3 = [(3 / 2)3]1/3

x3/3 = (3 / 2)3/3

x1 = (3 / 2)1

x = 3 / 2.

answered Mar 20, 2013 by diane Scholar

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