Quadratic Equations

Daisuke-kun SG

a) the roots of the equation 3x^2 +kx + 96 =0 are both positive and one is twice as large as the other Calculate the value of each root and find k

b)Given taht p^2 +q^2 = 13 and that pq =16, constructthe quadratic equation whose roots are p^2 and p^2. Find all possible values of p.

just learnt so not such.

help pls

^tamago^ SG

3x²+kx+96=0

For roots to be real and distinct, b² - 4ac > 0
k² - 4(3)(96) > 0
k² > 1152
k ≥ 34 or k ≤ -34 (k ∈ Z)

k = -3√(k²-1152)
8k² = 10368
k² = 1296
k = ± 36

for k = 36, x²+12x+32=0
(x+8)(x+4) = 0
x = -4 and -8 (rej. as x > 0)

for k = -36, x²-12x+32=0
(x-8)(x-4) = 0
x = 4 and 8

∴ The two roots are 4 and 8, and the value of k is -36.

Anyway using (x-a)(x-2a) should be an easier method as depicted below.

p²+q²=13 ⇒ p²=13-q² ⇒ q²=13-p²
pq=16

(x-p²)(x-q²) = 0
x²-(p²+q²)x+p²q² = 0
x²-13x+256 = 0

p²+q²=13 −−− (1)
pq=16 −−− (2)
sub (2) into (1): p²+(16/p)²=13
(p²)²-13p²+256=0
sub x into p², we have:
x²-13x+256 = (x-p²)(x-q²) = 0
⇒ p² = q² or p² = p² (rej.)
sub p²=q² into (1): 2p² = 13
p² = 6.5
p = ±2.55 (3 s.f.)

deathscythe99 SG

Hmmm for first question, u work backwards.

Assume a is the first root, 2a is the second root (twice as large)

Therefore, (x-2a) (x-a) = 0 is the equation for the roots.

x^2 - 3ax + 2a^2 = 0
3x^2 - 9ax + 6a^2 = 0

Comparing with 3x^2 + kx + 96 = 0,

6a^2 = 96, thus...a is +4 or -4
Solving for k, k can be -36 or +36 respectively.

Dun understand second question. Typo error?

davidche SG

Wow all good ways.

But tamago, is

''For roots to be real and distinct, b² - 4ac > 0
k² - 4(3)(96) > 0
k² > 1152
k ≥ 34 or k ≤ -34 (k ∈ Z) ''

needed to get answer?

deathscythe99 SG

Wow all good ways.

But tamago, is

''For roots to be real and distinct, b² - 4ac > 0
k² - 4(3)(96) > 0
k² > 1152
k ≥ 34 or k ≤ -34 (k ∈ Z) ''

needed to get answer?

Tamago's right. U need to prove which is the correct pair of answers. I didn't do that, so I get half the marks =X

davidche SG

second qn looks like must use alpha beta lol

^tamago^ SG

Wow all good ways.

But tamago, is

''For roots to be real and distinct, b² - 4ac > 0
k² - 4(3)(96) > 0
k² > 1152
k ≥ 34 or k ≤ -34 (k ∈ Z) ''

needed to get answer?

u're right, not needed. it says distinct so thought just do a quick one to acertain roughly what is k.