Help, maths! (1 Viewer)

[Frostbitten]

If 5x^2 - 37x + 68 =(meant to be 3 parallel lines) a(x-4)^2 + b(x-4) + c

i) Find the values of a, b and c
ii) Hence, factorise 5x^2 - 37x + 68

 

[cineti970128]

i will help you dw
just a minute that's all it takes

 

[cineti970128]

RHS = ax^2 + (b-8a)x + (16a - 4b +c)

if you learned simultaneous equations and equating coeffcients technique
no prob.
assuming you do not know
if
ax^2 + bx + c = px^2 + qs + r
p=a
q=b
r=c
hence using this and simultaneous equations
a= 5
b= 3
c=0

 

[Drongoski]

One way to do it:

Since the 2 are identically equal: it holds true for any value of x. So make 3 simple & smart substitutions.

x=4: you end up with c = 0 straight away.

x=0 gives you: 16a-4b = 68 or 4a-b = 17

x=1 gives you: 9a-3b = 12 or 3a-b = 12

Then solve the 2 simple simultaneous equations.

 

[Frostbitten]

RHS = ax^2 + (b-8a)x + (16a - 4b +c)

if you learned simultaneous equations and equating coeffcients technique
no prob.
assuming you do not know
if
ax^2 + bx + c = px^2 + qs + r
p=a
q=b
r=c
hence using this and simultaneous equations
a= 5
b= 3
c=0

Would you happen to know how to do the (ii) part?

 

[Drongoski]

Would you happen to know how to do the (ii) part?

The factorisation is shown done in an easy fashion in my 1st line! Maybe you cannot follow the way I did it

 

[Frostbitten]

Also, if anyone is keen.
For what values of k does the quadratic equation kx^2 + (k+3)x - 1 = 0 have real roots?

 

[GoldyOrNugget]

real roots exist when the discriminent b^2-4ac >=0 i.e. (k+3)^2 - 4(k)(-1) >= 0.

 

[Frostbitten]

The factorisation is shown done in an easy fashion in my 1st line! Maybe you cannot follow the way I did it

Yep you completely lost me :/

 

[Frostbitten]

real roots exist when the discriminent b^2-4ac >=0 i.e. (k+3)^2 - 4(k)(-1) >= 0.

Yeh, I got left with a quadratic k^2 + 10x + 9 > 0

i.e. (k + 9)(k+1)>0
k=-9, k=-1

Do I have to then test those values or?

 

[GoldyOrNugget]

You can test values in each interval, or you can picture the parabola that is concave down (because the coefficient of x^2 is positive) with x-intercepts at x=-9 and x=-1 -- for what x values is the parabola > 0 i.e. above the x-axis?

 

[Drongoski]

Yep you completely lost me :/

5(x-4)^2 + 3(x-4) means you have 5(x-4) of (x-4) plus 3 of (x-4)

therefore you have altogether [5(x-4) plus 3] of (x-4)

i.e. [5x-20 + 3] of (x-4)

i.e. [5x-17] of (x-4)

In other words you have: (5x-17) . (x-4)

 

[Frostbitten]

5(x-4)^2 + 3(x-4) means you have 5(x-4) of (x-4) plus 3 of (x-4)

therefore you have altogether [5(x-4) plus 3] of (x-4)

i.e. [5x-20 + 3] of (x-4)

i.e. [5x-17] of (x-4)

In other words you have: (5x-17) . (x-4)

How exactly did you get the 3(x-4) part?

 

[Drongoski]

How exactly did you get the 3(x-4) part?

I had in my 1st post put the cart before the horse. What I had shown in the 1st line was after you have found the values of a, b and c to be a=5, b=3 and c=0. So the quadratic in x became the quadratic in (x-4), viz: 5(x-4)^2 + 3(x-4) + 0

I hope this clears up any remaining confusion.

 

[cricketfan1997]

a(x^2-8x+16)+bx-4b+c
ax^2 - 8ax + 16a + bx - 4b + c
ax^2 - (8a+b)x + 16a - 4b + c
so a = 5
-(8 x 5) = b = -37
so b = 3
15(5) - 4(3) +c = 68
so c = 0

Hence, 5(x-4)^2+ 3(x-4) + 0
(x-4) (5(x-4) + 3)
= (x-4)(5x-17)

Q2. b^2 - 4ac
k^2 + 6k + 9 + 4k >= 0
k^2 + 10k + 9 >= 0
so it is real when k >= -1, and when k<= -9