TS EAMCET · Maths · Quadratic Equation

The sum of two roots of the equation $x^4-x^3-16 x^2+4 x+48=0$ is zero. If $\alpha, \beta, \gamma, \delta$ are the roots of this equation, then $\alpha^4+\beta^4+\gamma^4+\delta^4=$

A 123
B 369
C 132
D 396

Verified Solution

Answer & Solution

Correct Answer

(B) 369

Step-by-step Solution

Detailed explanation

$\alpha+\beta=0 \Rightarrow \beta=-\alpha$…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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$\log _4 2-\log _8 2+\log _{16} 2-\ldots$ is equal to

TS EAMCET 2010 Medium

Match the following.

	Column-1 Reaction		Column-2 Main product
A	${SO}_{3} \overset{heavy water}{⟶}$	I	$C_{2} D_{2}$
B	${CaC}_{2} \overset{heavy water (excess)}{⟶}$	II	${CD}_{4}$
C	${Al}_{4} C_{3} \overset{heavy water (large excess)}{⟶}$	III	$D_{2} {SO}_{3}$
		IV	$C_{2} D_{4}$
		V	$D_{2} {SO}_{4}$

The correct match is

TS EAMCET 2021 Hard

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The sum of two roots of the equation \(x^4-x^3-16 x^2+4 x+48=0\) is zero. If \(\alpha, \beta, \gamma, \delta\) are the roots of this equation, then \(\alpha^4+\beta^4+\gamma^4+\delta^4=\)

Answer & Solution

Step-by-step Solution

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