math doubts [NEW]

[Ghost Of Highbury]

how sir?

9 - (x-3)² is correct right?

(i edited it now)

(x+3)^2 - 9 doesnt give 6x - x^2

[slvri]

18/x⁴ + 1/x² =4
multiply by x⁴
18+x²=4x⁴
4x⁴-x²-18=0
4(x²)²-x²-18=0
u can see that this is a quadratic equation in x²
factorising we have
4x⁴-9x²+8x²-18=0
x²(4x²-9)+2(4x²-9)=0
4(x²-9)(x²+2)=0
either 4x²-9=0 or x²+2=0
4x²=9 or x²=-2
x²=9/4 (x²=-2 has no real roots, complex roots are +-sqrt(2)i)
so from the first one
x=sqrt(9/4)
x=+-(3/2)





the ones in red i didnt get it....plz elaborate...
y is 4x²-9=0? wont it be x²-9=0?

sory i made a mistake there
and for the first one, u multiply each term in the equation by x^4 to turn it into a expression that u can factorise easily

[zara]

sory i made a mistake there
and for the first one, u multiply each term in the equation by x^4 to turn it into a expression that u can factorise easily

yea Thanks...adi expalined it...=)

[slvri]

yea Thanks...adi expalined it...=)

i hope u understood

[zara]

yes slvri i did!

okay one more ques...

function f and g are defined by:
f(x): k-x    k is a constant
g(x): 9/x+2   whr x nt = 2   both x E R

Find the values of k for whch the eqtn f(x)=g(x) has two equal roots and solve the equation f(x)=g(x) in these cases.

[slvri]

yes slvri i did!

okay one more ques...

function f and g are defined by:
f(x): k-x    k is a constant
g(x): 9/x+2   whr x nt = 2   both x E R

Find the values of k for whch the eqtn f(x)=g(x) has two equal roots and solve the equation f(x)=g(x) in these cases.

f(x)=g(x)
k-x=9/x+2
(k-x)(x+2)=9
kx-x²+2k-2x=9
x²+(2-k)x+9-2k=0
in this equation a=1, b=2-k, c=9-2k
the condition for the equation f(x)=g(x) to have two equal roots is b²-4ac=0
(2-k)²-4(1)(9-2k)=0
4-4k+k²-36+8k=0
k²+4k-32=0
k²+8k-4k-32=0
k(k+8)-4(k+8)=0
(k+8)(k-4)=0
k=-8 or k=4
when k=-8
x²+(2-k)x+9-2k=0
x²+(2--8)x+9-2(-8)=0
x²+10x+25=0
(x+5)²=0
x+5=0
x=-5
but when k=4
x²+(2-k)x+9-2k=0
x²+(2-4)x+9-2(4)=0
x²-2x+1=0
(x-1)²=0
x-1=0
x=1
there u go

[zara]

function f and g are defined as

f(x): x²-2x
g(x):2x+3     both x E R


1) find the set of values of x for which f(x)>15
2)find the range of f...

im bad at this...had nly 2 or 3 classes of it...n didnt get much

[Ghost Of Highbury]

i guess the 1st one is -3>x>5

[astarmathsandphysics]

function f and g are defined as

f(x): x²-2x
g(x):2x+3     both x E R


1) find the set of values of x for which f(x)>15
2)find the range of f...

im bad at this...had nly 2 or 3 classes of it...n didnt get much

so

Complete the square

so so or so

[slvri]

function f and g are defined as

f(x): x²-2x
g(x):2x+3     both x E R


1) find the set of values of x for which f(x)>15
2)find the range of f...

im bad at this...had nly 2 or 3 classes of it...n didnt get much

its ok with practice u will get better
1) f(x)>15
x²-2x>15
x²-2x-15>0
x²-5x+3x-15>0
x(x-5)+3(x-5)>0
(x-5)(x+3)>0
(x-5))(x-(-3))>0
now theres a rule u need to keep in mind (i cant illustrate it here using a diagram so u will have to memorise it for the time being)
whenever an expression of the form(x-a)(x-b)>0 occurs where a<b then the solution is x<a and x>b
over here a is 5 and b is -3(5 is greater than -3)
so the solution is x<-3 and x>5
2)x²-2x
=(x²-2x+1)-1
=(x-1)²-1
now u can see that when x=1
f(1)=(1-1)²-1=-1
see for yourself that whatever value u put for x, the answer will always be greater than -1(-1 is the minimum value of f(x))
so the range of f(x) is f(x)>=-1

[Ghost Of Highbury]

shudnt -3>x>5

[slvri]

Hw cnw b greater than 5 and less at the same time?

[slvri]

I meant to say hw cn x b greater than 5 and less than -3 at the same time?

[Ghost Of Highbury]

yeah..then..how can it be less than -5 and greater than 3 at the same time??

[nid404]

Should be (x+3)²-9

yeah it should be