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Apr
28

Applications in practical situations

Posted by AskTuition at 9:00 pm

Linda bought 6 bars of chocolate and 8 cans of soft drinks for $12.60.Cindy bought 4 cans  of soft drinks  and 6 packets of crackers for $9.90.Joanne bought 1 bar of choclate,4 cans of drink and 4 packets of crackers.How much did Joanne pay altogether?

 

Suggested Solution

Let the number of chocolate bars be X.

Let the number of cans of soft drinks be Y.

Let the number of packets of crackers be Z.

6X + 8Y = 12.60 – eqn (1)

4Y + 6Z = 9.90 – eqn (2)

Find X + 4Y + 4Z?

4 * eqn(2) :   16Y + 24 Z = 39.60  - eqn (3)

eqn (3) + eqn (1):     6X + 24Y + 24Z = 52.20

Simplifying the above equation:  X +Y+Z= 8.70

Joanne paid $8.70 altogether.

 

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Mar
31

Polynomials

Posted by AskTuition at 9:50 pm

The roots alpha and beta of the equation a*x^2 + b*x + c = 0 are in the ratio 1:n.
Show that (1 + n)^2 *a*c= n*b^2

Suggested Solution

Given ratio is 1:n.

Hence, beta= n(alpha) such that alpha/beta = 1/n

Sum of roots:

alpha+n(alpha) = -b/a
=> (n+1)alpha = -b/a   – eqn 1
product of roots:
alpha(n*alpha)= n*alpha^2=c/a   –  eqn 2
[eqn(1)]^2 / eqn (2):
(n+1)^2 / n^2 = b^2/(ac)
Rearranging: (n+1)^2*ac = nb^2 (shown)
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Mar
31

Polynomials

Posted by AskTuition at 7:58 pm

The equation 2x^2 – x – 2 = 0 has roots α and β. Find the values of α^4 + β^4

Suggested Solution

We have a=2, b=-1 and c=-2

Using the quadratic formula,

We have

(-(-1) ± √ [(-1)2-4*(2)(-2)] )/ (2*2)

=(1 ± √ 17)/4

α4 + β4

=[(1 + √ 17)/4] 4 + [(1 - √ 17)/4] 4

=[(18 + 2√ 17)/16] 2 + [(18 -2 √ 17)/16] 2

=[(18 + 2√ 17)/16] 2 + [(18 -2 √ 17)/16] 2

=[(392 + 4√ 17)/256] + [(392 - 4√ 17)/256]

= 49/16

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Feb
28

Polynomials

Posted by AskTuition at 1:29 pm

If the roots of the equation x^2 + 3x + 4 = 0 are alpha and beta, find the value of alpha – beta.

Suggested Solution

Using the quadratic formula, the roots of the equation

x^2 + 3x + 4 is

[-3 + sq (-7)]/2 (alpha) and  [-3 - sq (-7)]/2 (beta)

Taking [-3 + sq (-7)]/2 minus [-3 - sq (-7)]/2,

We have sq (-7), which is the value of alpha – beta.

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Feb
28

Polynomials

Posted by AskTuition at 1:02 pm

If alpha and beta are the roots of the equation 5x^2 – 3x – 1 = 0, find the equation whose roots are 1/alpha^2 and 1/beta^2.

Suggested Solution

For the roots alpha and beta quadratic equations:

alpha + beta = – b/a

alpha*beta = c/a

In this case

a + b = 3/5

ab = -1/5

We can write an equation if we know the sum and the product of it’s roots:

x^2 – Sx + P = 0

the sum of the roots, S, we need is

1/a^2 + 1/b^2 =(a^2 + b^2) / (ab)^2

= [(a +b)^2 – 2ab] / (ab)^2

= [(3/5)^2-2(-1/5)]/(-1/5)^2

= (9/25 + 2/5)/ (1/25)

=19

P = (1/a^2) *( 1/b^2) = 1/(ab)^2=25

Hence the equation is

x^2 -19x + 25 = 0

 

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Feb
20

Polynomials

Posted by AskTuition at 2:33 pm

Given that a and b are the roots of the equation 3x^2+4x-2 = 0, form equations whose roots are:
a)2a, 2b

b) a/b, b/a

Suggested Solution

a)

(x-a)(x-b) =  3x^2+4x-2

For the new equation:

(x-2a)(x-2b) = 4(x/2-a)(x/2-b)

If (x-a)(x-b) =  3x^2+4x-2

(x/2-a)(x/2-b) = 3(x/2)^2+4(x/2)-2

= ¾ x^2 + 2x – 2

4(x/2-a)(x/2-b) =3 x^2 + 8x – 8

b)

For the roots alpha and beta quadratic equations:

alpha + beta = – b/a

alpha*beta = c/a

In this case

a + b = -4/3

ab = -2/3

We can write an equation if we know the sum and the product of it’s roots:

x^2 – Sx + P = 0

the sum of the roots, S, we need is

a/b + b/a =(a^2 + b^2) / ab

= [(a +b)^2 – 2ab] / ab

= [(-4/3)^2-2(-2/3)]/(-2/3)

= (16/9 + 4/3)/ (-2/3)

=28/9 * (-3/2)

= -14/3

P = a/b * b/a = 1

Hence the equation is

x^2 – (-14/3)x + 1 = 0

x^2 + 14/3x + 1 = 0

 

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