Q1.
& are real roots of above quadratic
& is integer,
Find value of .
Solution:
Q2.
, roots for this quadratic are & .
Find value of
α2+2α(−5−2α)=−234\alpha^2 + 2\alpha(-5 - 2\alpha) = -\frac{23}{4}
α2−10α−4α2=−234\alpha^2 - 10\alpha - 4\alpha^2 = -\frac{23}{4}
−10α−3α2=−234-10\alpha - 3\alpha^2 = -\frac{23}{4}
12α2+40α=2312\alpha^2 + 40\alpha = 23
12α2+40α−23=012\alpha^2 + 40\alpha - 23 = 0
12α2−6α+46α−23=012\alpha^2 - 6\alpha + 46\alpha - 23 = 0
6α(2α−1)+23(2α−1)=06\alpha(2\alpha - 1) + 23(2\alpha - 1) = 0
(6α+23)(2α−1)=0(6\alpha + 23)(2\alpha - 1) = 0
α=−236orα=12\alpha = -\frac{23}{6} \quad \text{or} \quad \boxed{\alpha = \frac{1}{2}}
β=−5−2α\beta = -5 - 2\alpha
β=−5−2(−236)⇒β=−5+233⇒β=83\beta = -5 - 2\left(-\frac{23}{6}\right)
\Rightarrow \beta = -5 + \frac{23}{3}
\Rightarrow \beta = \frac{8}{3}
(Alternative branch):
β=−5−2⋅12=−6\beta = -5 - 2 \cdot \frac{1}{2} = -6
α2β=−32\alpha^2 \beta = -\frac{3}{2}
α=12,β=−6\alpha = \frac{1}{2}, \quad \beta = -6
Roots are
12,12,−6\boxed{\frac{1}{2}, \frac{1}{2}, -6}
Q4.
Evaluate the expression:
(α+β+γ)(α+β+δ)(α+γ+δ)(β+γ+δ)(\alpha + \beta + \gamma)(\alpha + \beta + \delta)(\alpha + \gamma + \delta)(\beta + \gamma + \delta)
Given the polynomial:
x4−7x2+x−5=0x^4 - 7x^2 + x - 5 = 0
with roots α,β,γ,δ\alpha, \beta, \gamma, \delta.
Step-by-step breakdown:
-
From Vieta's formulas, we know for a polynomial
x4+bx3+cx2+dx+e=0x^4 + bx^3 + cx^2 + dx + e = 0, the sum of the roots is:
α+β+γ+δ=−ba\alpha + \beta + \gamma + \delta = -\frac{b}{a}
For our polynomial, a=1a = 1 and b=0b = 0, so:
α+β+γ+δ=0\alpha + \beta + \gamma + \delta = 0
-
Using this, we can rewrite each term:
-
α+β+γ=−δ\alpha + \beta + \gamma = -\delta
-
α+β+δ=−γ\alpha + \beta + \delta = -\gamma
-
α+γ+δ=−β\alpha + \gamma + \delta = -\beta
-
β+γ+δ=−α\beta + \gamma + \delta = -\alpha
So the product becomes:
(−δ)(−γ)(−β)(−α)=αβγδ(-\delta)(-\gamma)(-\beta)(-\alpha) = \alpha\beta\gamma\delta
-
From Vieta’s formula again:
αβγδ=ea=−51=−5\alpha\beta\gamma\delta = \frac{e}{a} = \frac{-5}{1} = -5
✅ Final Answer:
(α+β+γ)(α+β+δ)(α+γ+δ)(β+γ+δ)=−5
Q5.
x³ - 4x² + 7x - 4 = 0
α, β, γ are roots
Evaluate:
a) α² + β² + γ²
b) 1/α + 1/β + 1/γ
S₁ = 4
S₂ = 7
S₃ = 4
a) (α + β + γ)² = α² + β² + γ² + 2(αβ + βγ + γα)
S₁² = α² + β² + γ² + 2S₂
α² + β² + γ² = S₁² - 2S₂
= 4² - 2(7)
= (16 - 14)
= 2
b)
1/α + 1/β + 1/γ = (αβ + βγ + γα) / (αβγ)
= S₂ / S₃
= 7 / 4
Q6.
Here is the handwritten content converted to text:
VI) If α & β are two real nos. satisfying
2α² - 3α + 1 = 0
& 2β² - 3β + 1 = 0.
Find possible values of
α/β + β/α .
Solution:
There are 2 cases here:
1)
α & β are same => answer: 2
2) α & β are different => they are roots of this quadratic: 2.x^2 - 3x + 1 = 0
Solving that for roots you will get: 5/2
Q7.
Given that p + q = 198
Find integer solution of
x² + px + q = 0
Solution:
___________________________
x² + px + q = 0 → α, β
α + β = -p
αβ = q
p + q = 198
-α - β + αβ = 198
1 -α - β + αβ = 199
(1 - α)(1 - β) = 199
α β
0 -198
-198 0
2 200
200 2
(1 - α)(1 - β) = 199
→ Possible factor pairs:
1 × 199
199 × 1
-1 × -199
-199 × -1
Boxed result:
either 0, -198
or 2, 200
Note:
n is prime
← 14
2, 3, 5, 7, 11, 13
Theory Note:
z = α + iβ is root of any polynomial
then z̄ = α - iβ is also the root of polynomial
→ Imaginary root always exist in pair.
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