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3 years ago

The difference between the roots of the equation 2x2−5x+c=0 is 0.25. Find c.

Mathematics

1 answer:

vredina [299]3 years ago

7 0

Answer:

c = 3.09375

Step-by-step explanation:

let the roots be α and β

2x2−5x+c=0

Comparing withe the standard equation ax² + bx + c = 0

a=2 b = -5 and c=c

sum : α + β = $\frac{-b}{a}$

product : αβ = $\frac{c}{a}$

α + β = $\frac{--5}{2}$ = $\frac{5}{2}$ = 2.5 -------------------------------------(1)

αβ = $\frac{c}{2}$ -----------------------------------------------------------(2)

the question says "the difference between the roots of the equation is 0.25"

so, α -β = 0.25 ---------------------------------------------------(3)

we can use equation (1) and equation (3) to find the values of our α and β by solving simultaneously

add equation (1) and equation (3)

2 α = 2.75

α = 2.75 /2 = 1.375

subtract equation (3) from equation (1)

2β = 2.25

β = 2.25 /2 = 1.125

we can now use equation (2) to find the value of c

αβ = $\frac{c}{2}$

1.375 × 1.125 = $\frac{c}{2}$

1.546875 = $\frac{c}{2}$

multiply both-side of the equation by 2

1.546875 × 2 = c

3.09375 = c

c = 3.09375

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Central Boulevard runs diagonally across town as shown below. The distance from Eastern Avenue to Western Avenue along South Str

murzikaleks [220]

The locations of the streets describes a right triangle with the angles 30 degrees and 60 degrees between them and the length of the side opposite the 60 degrees angle is $4\sqrt{3} \, miles$

a.) To find the distance form North Street to South Street along Eastern Avenue, we find the length of the side opposite the 30 degrees angle.

Using sine rule, we recall that the sine rule states that
$\frac{\sin{A}}{a} = \frac{\sin{B}}{b}$
where:
a = $4\sqrt{3}$
A = 60 degrees
B = 30 degrees
b = unknown
$\frac{\sin{60^o}}{4\sqrt{3}} = \frac{\sin{30^o}}{b} \\ b=\frac{4\sqrt{3} \sin{30^o}}{\sin{60^o}}= \frac{3.4641}{0.8660} = 4 \, miles$

Therefore, the distance form North Street to South Street along Eastern Avenue is 4 miles.

b.) To find the distance from North Street to South Street along Central Boulevard we find the length of the hypothenus of the right triangle.

Using pythagoras theorem, we recall that the pythagoras states that
$c^2=a^2+b^2$
where:
a = $4\sqrt{3}$
b = 4 miles
c = hypothenus = unknown
$c^2=\left(4 \sqrt{3} \right)^2+4^2=48+16=64 \\ c= \sqrt{64} =8 \, miles$

Therefore, the distance from North Street to South Street along Central Boulevard is 8 miles

5 0

3 years ago

What am I doing right

Oduvanchick [21]

Although your -3 point is correct you need to graph the other point on (3,1) rather than (3,0)
so change that.

5 0

3 years ago

Expand and simplify

Blababa [14]

<h2>Question:</h2>

→To simple the given equations