Algebra theory class 3 solved examples

Q1.

x2+(2a−1)x+a2=0x^2 + (2a - 1)x + a^2 = 0
α,β>0\alpha, \beta > 0

α\alpha & β\beta are real roots of above quadratic
& aa is integer,

Find value of ∣α−β∣|\sqrt{\alpha} - \sqrt{\beta}|.

Solution:

S1=α+β=−(2a−1)1S_1 = \alpha + \beta = -\frac{(2a - 1)}{1}
α+β=1−2a\alpha + \beta = 1 - 2a

S2=αβ=a21S_2 = \alpha \beta = \frac{a^2}{1}
αβ=a2\alpha \beta = a^2

(α−β)2=(α)2+(β)2−2αβ(\sqrt{\alpha} - \sqrt{\beta})^2 = (\sqrt{\alpha})^2 + (\sqrt{\beta})^2 - 2\sqrt{\alpha\beta} =α+β−2αβ= \alpha + \beta - 2\sqrt{\alpha\beta} =1−2a−2a2= 1 - 2a - 2\sqrt{a^2} =1−2a−2∣a∣= 1 - 2a - 2|a|

D≥0D \geq 0 ax2+bx+c=0ax^2 + bx + c = 0 1x2+(2a−1)x+a2=01x^2 + (2a - 1)x + a^2 = 0 b2−4ac≥0b^2 - 4ac \geq 0 (2a−1)2−4(1)(a2)≥0(2a - 1)^2 - 4(1)(a^2) \geq 0 4a2−4a+1−4a2≥04a^2 - 4a + 1 - 4a^2 \geq 0 1−4a≥01 - 4a \geq 0 1≥4a1 \geq 4a a≤14a \leq \frac{1}{4}

(α−β)2=1−2a−2∣a∣(\sqrt{\alpha} - \sqrt{\beta})^2 = 1 - 2a - 2|a| =1−2a−2(−a)= 1 - 2a - 2(-a) =1−2a+2a= 1 - 2a + 2a =1= 1 ∣α−β∣=1\boxed{|\sqrt{\alpha} - \sqrt{\beta}| = 1} ∣x∣={xif x>0−xif x<0|x| = \begin{cases} x & \text{if } x > 0 \\ -x & \text{if } x < 0 \end{cases}
Q2.

x2+x−3=0x^2 + x - 3 = 0, roots for this quadratic are α\alpha & β\beta.

Find value of α3−4β2+19.

Solution:\alpha^3 - 4\beta^2 + 19

α is root ⇒α2+α−3=0\alpha \text{ is root } \Rightarrow \alpha^2 + \alpha - 3 = 0 β is root ⇒β2+β−3=0\beta \text{ is root } \Rightarrow \beta^2 + \beta - 3 = 0 α2+α−3=0⇒α2=3−α⇒α3=3α−α2⇒α3=3α−(3−α)⇒α3=4α−3\alpha^2 + \alpha - 3 = 0 \Rightarrow \alpha^2 = 3 - \alpha \Rightarrow \alpha^3 = 3\alpha - \alpha^2 \Rightarrow \alpha^3 = 3\alpha - (3 - \alpha) \Rightarrow \boxed{\alpha^3 = 4\alpha - 3} β2=3−β\boxed{\beta^2 = 3 - \beta} α3−4β2+19=4α−3−4(3−β)+19\alpha^3 - 4\beta^2 + 19 = 4\alpha - 3 - 4(3 - \beta) + 19

=4α−3−12+4β+19= 4\alpha - 3 - 12 + 4\beta + 19 =4(α+β)+4= 4(\alpha + \beta) + 4 =4(−1)+4= 4(-1) + 4 =0


Q3.
4x3+20x2−23x+6=04x^3 + 20x^2 - 23x + 6 = 0

Two roots of cubic are equal, then find all roots.

Solution:

Let the roots be α,α,β\alpha, \alpha, \beta

Cubic equation:

ax3+bx2+cx+d=0ax^3 + bx^2 + cx + d = 0

Given:

4x3+20x2−23x+6=04x^3 + 20x^2 - 23x + 6 = 0

Using symmetric sums:

S1=α+α+β=−ba=−204⇒2α+β=−5S_1 = \alpha + \alpha + \beta = \frac{-b}{a} = \frac{-20}{4} \Rightarrow \boxed{2\alpha + \beta = -5} S2=α2+αβ+αβ=ca=−234⇒α2+2αβ=−234S_2 = \alpha^2 + \alpha\beta + \alpha\beta = \frac{c}{a} = \frac{-23}{4} \Rightarrow \boxed{\alpha^2 + 2\alpha\beta = -\frac{23}{4}} S3=α⋅α⋅β=−da=64⇒α2β=32S_3 = \alpha \cdot \alpha \cdot \beta = \frac{-d}{a} = \frac{6}{4} \Rightarrow \boxed{\alpha^2 \beta = \frac{3}{2}}

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:

  1. 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

  2. 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

  3. 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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