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3 years ago

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The equation 2x^2 + x - 1 = 0 has two solutions. Find an equation of the form ax^2 + bx + c = 0, which solutions....

1 answer:

leonid [27]3 years ago

4 0

Answer:

Step-by-step explanation:

Let the solution to

2x^2 + x -1 =0

x^2+ (1/2)x -(1/2)

are a and b

Hence a + b = -(1/2) ( minus the coefficient of x )

ab = -1/2 (the constant)

A. We want to have an equation where the roots are a +5 and b+5.

Therefore the sum of the roots is (a+5) + (b+5) = a+ b +10 =(-1/2) + 10 =19/2.

The product is (a+5)(b+5) =ab + 5(a+b) + 25 = (-1/2) + 5(-1/2) + 25 = 22.

So the equation is

x^2-(19/2)x + 22 =0

2x^2-19x + 44 =0

B. We want the roots to be 3a and 3b.

Hence (3a) + (3b) = 3(a+b) = 3(-1/2) =-3/2 and

(3a)(3b) = 9(ab) =9(-1/2)=-9/2.

So the equation is

x^2 +(3/2) x -9/2 = 0

2x^2 + 3x -9 =0.

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i f N(-7,-1) is a point on the terminal side of ∅ in standard form, find the exact values of the trigonometric functions of ∅.

Natali [406]

Answer:

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = -5\sqrt 2$

Step-by-step explanation:

Given

$N = (-7,-1)$ --- terminal side of $\varnothing$

Required

Determine the values of trigonometric functions of $\varnothing$ .

For $\varnothing$ , the trigonometry ratios are:

$sin\ \varnothing = \frac{y}{r}$ $cos\ \varnothing = \frac{x}{r}$ $tan\ \varnothing = \frac{y}{x}$

$cot\ \varnothing = \frac{x}{y}$ $sec\ \varnothing = \frac{r}{x}$ $csc\ \varnothing = \frac{r}{y}$

Where:

$r^2 = x^2 + y^2$

$r = \sqrt{x^2 + y^2$

In $N = (-7,-1)$

$x = -7$ and $y = -1$

So:

$r = \sqrt{(-7)^2 + (-1)^2$

$r = \sqrt{50$

$r = \sqrt{25 * 2$

$r = \sqrt{25} * \sqrt 2$

$r = 5 * \sqrt 2$

$r = 5 \sqrt 2$

<u>Solving the trigonometry functions</u>

$sin\ \varnothing = \frac{y}{r}$

$sin\ \varnothing = \frac{-1}{5\sqrt 2}$

Rationalize:

$sin\ \varnothing = \frac{-1}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$sin\ \varnothing = \frac{-\sqrt 2}{5*2}$

$sin\ \varnothing = \frac{-\sqrt 2}{10}$

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{x}{r}$

$cos\ \varnothing = \frac{-7}{5\sqrt 2}$

Rationalize

$cos\ \varnothing = \frac{-7}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$cos\ \varnothing = \frac{-7*\sqrt 2}{5*2}$

$cos\ \varnothing = \frac{-7\sqrt 2}{10}$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{y}{x}$

$tan\ \varnothing = \frac{-1}{-7}$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = \frac{x}{y}$

$cot\ \varnothing = \frac{-7}{-1}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{r}{x}$

$sec\ \varnothing = \frac{5\sqrt 2}{-7}$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = \frac{r}{y}$

$csc\ \varnothing = \frac{5\sqrt 2}{-1}$

$csc\ \varnothing = -5\sqrt 2$

3 0

3 years ago

28 = 4/5 x - 8 What value of x makes the equation true?

Harrizon [31]

Answer:

x will have to be 45 to make this true

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

(Algebra 2)<br><br> Find the inverse of f(x) and its domain.

elixir [45]

Let f(x)=y
f^-1(x)=x
y=1/(x+5) - 1
y+1=1/(x+5)
(x+5)=1/(y+1)
x=1/(y+1)-5
f^-1(x)=1/(x+1) -5

denominator cannot equals to zero
x+1≠0
x≠-1

answer
B

4 0

2 years ago

PLEASE HELP. I WILL GIVE BRAINLESS

Mila [183]

Answer:

1. 47.1% 2. 69.4% 3. 72.9% 4. 55.6%

Step-by-step explanation:

you just have to add both the numbers and divide the one you are trying to find is liked by the total of both.

ex.

what precent of chicago likes rain?

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143+59=202

143/202=70.7%

7 0

3 years ago

sukhopar [10]

Answer:

(a) 1km

(b)80 fence panels

Step-by-step explanation:

(a) perimeter = 300+300= 600

200+200= + 400

=1000m

1km = 1000m

(b).

1panel = 2.5 m 1× 200÷2.5 = 80

=200m

7 0

3 years ago