All categories

3 years ago

Suppose the roots of the equation 2x^2−5x−6=0 are α and β.Find the quadratic equation with roots 1/α and 1/β.

Mathematics

1 answer:

pochemuha3 years ago

8 0

Answer:

6x² + 5x - 2 = 0

Step-by-step explanation:

Given

2x² - 5x - 6 = 0 ← in standard form

with a = 2, b = - 5, c = - 6, then

sum of roots α + β = - $\frac{b}{a}$ = $\frac{5}{2}$

product of roots = $\frac{c}{a}$ = - 3, then

sum of new roots = $\frac{1}{\alpha }$ + $\frac{1}{\beta }$

= $\frac{\beta+\alpha }{\alpha \beta }$

= $\frac{\frac{5}{2} }{-3}$ = - $\frac{5}{6}$

product of new roots = $\frac{1}{\alpha }$ × $\frac{1}{\beta }$

= $\frac{1}{\alpha\beta }$ = - $\frac{1}{3}$

Hence the required equation is

x² + $\frac{5}{6}$ x - $\frac{1}{3}$ = 0 or

6x² + 5x - 2 = 0 ( multiplying through by 6 )

You might be interested in

Sin4x.sin5x+sin4x.sin3x-sin2x.sinx=0

andreev551 [17]

Recall the angle sum identity for cosine:

cos(x + y) = cos(x) cos(y) - sin(x) sin(y)

cos(x - y) = cos(x) cos(y) + sin(x) sin(y)

==> sin(x) sin(y) = 1/2 (cos(x - y) - cos(x + y))

Then rewrite the equation as

sin(4x) sin(5x) + sin(4x) sin(3x) - sin(2x) sin(x) = 0

1/2 (cos(-x) - cos(9x)) + 1/2 (cos(x) - cos(7x)) - 1/2 (cos(x) - cos(3x)) = 0

1/2 (cos(9x) - cos(x)) + 1/2 (cos(7x) - cos(3x)) = 0

sin(5x) sin(-4x) + sin(5x) sin(-2x) = 0

-sin(5x) (sin(4x) + sin(2x)) = 0

sin(5x) (sin(4x) + sin(2x)) = 0

Recall the double angle identity for sine:

sin(2x) = 2 sin(x) cos(x)

Rewrite the equation again as

sin(5x) (2 sin(2x) cos(2x) + sin(2x)) = 0

sin(5x) sin(2x) (2 cos(2x) + 1) = 0

sin(5x) = 0 or sin(2x) = 0 or 2 cos(2x) + 1 = 0

sin(5x) = 0 or sin(2x) = 0 or cos(2x) = -1/2

sin(5x) = 0 ==> 5x = arcsin(0) + 2nπ or 5x = arcsin(0) + π + 2nπ

… … … … … ==> 5x = 2nπ or 5x = (2n + 1)π

… … … … … ==> x = 2nπ/5 or x = (2n + 1)π/5

sin(2x) = 0 ==> 2x = arcsin(0) + 2nπ or 2x = arcsin(0) + π + 2nπ

… … … … … ==> 2x = 2nπ or 2x = (2n + 1)π

… … … … … ==> x = nπ or x = (2n + 1)π/2

cos(2x) = -1/2 ==> 2x = arccos(-1/2) + 2nπ or 2x = -arccos(-1/2) + 2nπ

… … … … … … ==> 2x = 2π/3 + 2nπ or 2x = -2π/3 + 2nπ

… … … … … … ==> x = π/3 + nπ or x = -π/3 + nπ

(where n is any integer)

5 0

3 years ago

I need help with letter B, please!

Gala2k [10]

So if you want to fit the y-intercepts or "b", on the y-axis you should go by 25's [0, 25, 50, 75, 100...]

If the x-axis does not have to follow the same pattern (25's), you should go by 5's [0, 5, 10, 15, 20...]

y = 7x + 50

y = 2x + 175

First I would plot the y-intercepts for each equation, then plot a few points with x = 5, 10, 15 Then draw a straight line.

The point where the two lines meet/cross paths is your solution. It should be (25, 225) The x-axis is the number of miles, and the y-axis is the total cost. So Truck driver A and B would arrive/be at the same place/meet in 25 miles at the same cost of $225