Question

If the roots of
x4 - 10x? +37x2 - 60x +36=0 are
a,a,ß,ß,(a<b), then 2a +3B-203=
|EAM 18 TS]
C​

Answer 1

EXPLANATION.

Roots : x⁴ - 10x³ + 37x² - 60x + 36 = 0.

⇒ α, α, β, β (α < β).

As we know that,

Sum of the zeroes of the biquadratic polynomial.

⇒ α + α + β + β = - b/a.

⇒ α + α + β + β = -(-10)/1 = 10.

⇒ 2α + 2β = 10.

⇒ α + β = 5. - - - - - (1).

⇒ α = 5 - β. - - - - - (1).

Products of the zeroes of the biquadratic polynomial.

⇒ (α)(α)(β)(β) = e/a.

⇒ α x α x β x β = 36.

⇒ α²β² = 36.

⇒ (αβ)² = (6)².

⇒ αβ = 6.

Put the values of α = 5 - β in the equation, we get.

⇒ (5 - β)(β) = 6.

⇒ 5β - β² = 6.

⇒ 6 + β² - 5β = 0.

⇒ β² - 5β + 6 = 0.

Factorizes the equation into middle term splits, we get.

⇒ β² - 3β - 2β + 6 = 0.

⇒ β(β - 3) - 2(β - 3) = 0.

⇒ (β - 2)(β - 3) = 0.

⇒ β = 2   and   β = 3.

Put the values of β in the equation (1), we get.

⇒ α = 5 - β.

Put the value of β = 2 in the equation, we get.

⇒ α = 5 - 2.

⇒ α = 3.

Put the value of β = 3 in the equation, we get.

⇒ α = 5 - 3.

⇒ α = 2.

⇒ α = 2  and  β = 3.

⇒ α = 3  and  β = 2.

It is clearly given that,

⇒ (α < β).

So, we take.

⇒ α = 2  and  β = 3.

To find :

⇒ 2α + 3β - 2αβ.

Put the values in the equation, we get.

⇒ 2(2) + 3(3) - 2(2)(3).

⇒ 4 + 9 - 12.

⇒ 13 - 12 = 1.

⇒ 2α + 3β - 2αβ = 1.

                                                                                                                   

MORE INFORMATION.

(1) = If D₁ + D₂ ≥ 0 : At least one of the equation has real roots.

(2) = If D₁ + D₂ < 0 : At least one of the equation has imaginary roots.

(3) = If D₁. D₂ < 0 : One equation has real and distinct roots and other has imaginary roots.

Answer 2

Answer:

→ Given, the quartic polynomial

$\rm \: {x}^{4} - 10 {x}^{3} + 37 {x}^{2} - 60x + 36 = 0$

And its zeroes are :

$\alpha , \alpha , \beta , \beta \: \: \sf \: \: \: \: where \: \: \: \alpha < \beta$

→ We have to find :

⇨ The value of :

1) [ I think + seeing others answer ! ]

$2 \alpha + 3 \beta - 2 \alpha \beta$

or

2) [according to the question ]

$2 \alpha + 3 \beta - 203$

Solution :

We know that,

$\rm \:If\:\:\: a {x}^{4} + b {x}^{3} + c {x}^{2} + dx + e \: \: is \: a \: \\ \sf \: quartic \: polynomial \: then, \:$

$\small{ \green{ \sf \: sum \: of \: zeroes \: of \: quartic \: polynomial \: = - \frac{coefficient \: of \: {x}^{3} }{coefficient \: of \: {x}^{4} } }}$

and

$\pink{ \sf \: product \: of \: zeroes \: = \frac{constant \: term}{coefficient \: of \: {x}^{4} } }$

According to the question,

Sum of zeroes :

$\orange{ \boxed{ \boxed{ \begin{array}{cc} \alpha + \alpha + \beta + \beta = - \frac{ - 10}{1} \\ \\ = > 2 \alpha + 2 \beta = 10 \\ \\ = > 2( \alpha + \beta ) = 10 \\ \\ = > \alpha + \beta = 5 \\ \\ = > \alpha = 5 - \beta \: \: \: - - - (1)\end{array}}}}$

and

product of zeroes :

$\orange{ \boxed{ \boxed{ \begin{array}{cc} \alpha \alpha \beta \beta = \frac{36}{1} \\ \\ = > { \alpha }^{2} { \beta }^{2} = 36 \\ \\ = > \alpha \beta = 6 \: \: \: \: - - - (2)\end{array}}}}$

Now, from eqn(1) and enq(2)

$\pink{ \boxed{ \boxed{ \begin{array}{cc} (5 - \beta ) \beta = 6 \\ \\ = > 5 \beta - { \beta }^{2} = 6 \\ \\ = > { \beta }^{2} - 5 \beta + 6 = 0 \\ \\ = > { \beta }^{2} - 3 \beta - 2 \beta + 6 = 0 \\ \\ = > \beta ( \beta - 3) - 2( \beta - 3) = 0 \\ \\ = > ( \beta - 3)( \beta - 2) = 0 \\ \\ \therefore \: \beta = 2 \: or \: 3\end{array}}}}$

putting the value of ß in eqn(1)

$\alpha = 5 - 2 = 3 \: \: \sf \: when \: \beta = 2 \\ \\ \bf \: and \\ \\ \alpha = 5 - 3 = 2 \: \: \: \sf \: when \: \beta = 3 \\ \\ \therefore \: \alpha = 2 \: or \: 3$

But in our condition,

$\alpha = 2 \: \: and \: \beta = 3 \: \: \{ \because \: \alpha < \beta \}$

Now,

1)

$2 \alpha + 3 \beta - 2 \alpha \beta \\ \\ = 2 \times 2 + 3 \times 3 - 2 \times 2 \times 3 \\ \\ = 4 + 9 - 12 \\ \\ = 13 - 12 \\ \\ = 1$

Or

2)

$2 \alpha + 3 \beta - 203 \\ \\ = 2 \times 2 + 3 \times 3 - 203 \\ \\ = 4 + 9 - 203 \\ \\ = - 190$