Simplify 9|2x-1|-2|6x-3|
Suggested Solution
9|2x-1|-2|6x-3|
= 9|2x-1|-6|2x-1|
= 3|2x-1|
= |6x-3|
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Factorise 6x^3 – 5x^2 – 8x + 3.
Hence solve 6x^3 – 5x^2 – 2x + 1 = 0
Suggested Solution
6x^3 – 5x^2 – 8x + 3
= (x+1)(6x^2 – 11x + 3)
= (x+1)(3x – 1)(2x – 3) (Factorised)
6x^3 – 5x^2 – 2x + 1 = 0
(6x^3 – 5x^2 – 8x + 3) + (6x – 2) = 0
(x+1)(3x – 1)(2x – 3) + 2(3x – 1) = 0
(3x – 1)(2x^2 – x – 1) = 0
(3x – 1)(2x + 1)(x – 1) =0
x=1/3 , -1/2 , 1
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.
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)
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
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.
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