Question

What values of b will cause 8x^2 x bx + 2 = 0 to have one real solution?

Answer 1

Answer:

To find the value of  

b

where there will be just ONE soluition, we set the discriminate equal to  

0

, substitute for  

a

and  

c

and solve for  

b

:

Substitute:

2

for  

a

−

b

for  

b

−

9

for  

c

−

b

2

−

(

4

⋅

2

⋅

−

9

)

=

0

−

b

2

−

(

−

72

)

=

0

−

b

2

+

72

=

0

−

b

2

+

72

−

72

=

0

−

72

−

b

2

+

0

=

−

72

−

b

2

=

−

72

−

1

⋅

−

b

2

=

−

1

⋅

−

72

b

2

=

72

√

b

2

=

±

√

72

b

=

±

√

36

⋅

2

b

=

±

√

36

√

2

b

=

±

6

√

2To find the value of  

b

where there will be just ONE soluition, we set the discriminate equal to  

0

, substitute for  

a

and  

c

and solve for  

b

:

Substitute:

2

for  

a

−

b

for  

b

−

9

for  

c

−

b

2

−

(

4

⋅

2

⋅

−

9

)

=

0

−

b

2

−

(

−

72

)

=

0

−

b

2

+

72

=

0

−

b

2

+

72

−

72

=

0

−

72

−

b

2

+

0

=

−

72

−

b

2

=

−

72

−

1

⋅

−

b

2

=

−

1

⋅

−

72

b

2

=

72

√

b

2

=

±

√

72

b

=

±

√

36

⋅

2

b

=

±

√

36

√

2

b

=

±

6

√

2

Step-by-step explanation:

Answer 2

Answer: b = 8 or b = -8

=======================================================

Explanation:

I'm assuming you meant to type 8x^2 + bx + 2 = 0

Compare this to ax^2 + bx + c = 0

We have a = 8, b = unknown, c = 2.

We'll use the discriminant formula which is D = b^2 - 4ac to find that

D = b^2 - 4ac

D = b^2 - 4*8*2

D = b^2 - 64

Since we want one real solution, we set the discriminant equal to zero.

D = b^2 - 64

0 = b^2 - 64

64 = b^2

b^2 = 64

b = sqrt(64) or b = -sqrt(8)

b = 8 or b = -8

So the equation 8x^2 + 8x + 2 = 0 has one solution. So does the equation 8x^2-8x+2 = 0