# Help, maths! (1 Viewer)

#### Frostbitten

##### Active Member
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

##### Member
just a minute that's all it takes

#### cineti970128

##### Member
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

##### Well-Known Member
$\bg_white 5x^2 - 37x + 68 \equiv 5(x-4)^2 +3(x-4) \equiv (x-4) [5(x-4) + 3] \equiv (x-4)(5x-17)$

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.

Last edited:

#### Frostbitten

##### Active Member
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

##### Well-Known Member
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

##### Active Member
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

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

#### Frostbitten

##### Active Member
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

##### Active Member
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

##### Señor Member
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

##### Well-Known Member
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

##### Active Member
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

##### Well-Known Member
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

##### Member
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