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Rzqust [24]
3 years ago
14

What values of b will cause 4x^2+bx+25=0 to have one real solution?

Mathematics
1 answer:
Darina [25.2K]3 years ago
5 0

Answer:

b=20 or b=-20

Step-by-step explanation:

Keep in mind the meanings of the values of the discriminant:

If b^2-4ac=0, then the quadratic will have only 1 solution (double root)

If b^2-4ac<0, then the quadratic will have no real solutions

If b^2-4ac>0, then the quadratic will have 2 unique solutions

In this case, to get one real solution, the discriminant must be set up as b^2-4ac=0. Setting up the equation we get:

b^2-4(4)(25)=0

b^2-400=0

b^2=400

b=20 or b=-20

So the values of b would have to be either 20 or -20

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Find two numbers between 100 and 150 with HCF 24
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Solve the following differential equations or initial value problems. In part (a), leave your answer in implicit form. For parts
shepuryov [24]

Answer:

(a) (y^5)/5 + y^4 = (t^3)/3 + 7t + C

(b) y = arctan(t(lnt - 1) + C)

(c) y = -1/ln|0.09(t + 1)²/t|

Step-by-step explanation:

(a) dy/dt = (t^2 + 7)/(y^4 - 4y^3)

Separate the variables

(y^4 - 4y^3)dy = (t^2 + 7)dt

Integrate both sides

(y^5)/5 + y^4 = (t^3)/3 + 7t + C

(b) dy/dt = (cos²y)lnt

Separate the variables

dy/cos²y = lnt dt

Integrate both sides

tany = t(lnt - 1) + C

y = arctan(t(lnt - 1) + C)

(c) (t² + t) dy/dt + y² = ty², y(1) = -1

(t² + t) dy/dt = ty² - y²

(t² + t) dy/dt = y²(t - 1)

(t² + t)/(t - 1)dy/dt = y²

Separating the variables

(t - 1)dt/(t² + t) = dy/y²

tdt/(t² + t) - dt/(t² + t) = dy/y²

dt/(t + 1) - dt/(t(t + 1)) = dy/y²

dt/(t + 1) - dt/t + dt/(t + 1) = dy/y²

Integrate both sides

ln(t + 1) - lnt + ln(t + 1) + lnC = -1/y

2ln(t + 1) - lnt + lnC = -1/y

ln|C(t + 1)²/t| = -1/y

y = -1/ln|C(t + 1)²/t|

Apply y(1) = -1

-1 = ln|C(1 + 1)²/1|

-1 = ln(4C)

4C = e^(-1)

C = (1/4)e^(-1) ≈ 0.09

y = -1/ln|0.09(t + 1)²/t|

8 0
3 years ago
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