TS EAMCET · Maths · Quadratic Equation
If \(\alpha, \beta, \gamma\) are the roots of \(x^3-3 x^2-4 x+12=0\), then \(\sum(\alpha+\beta)^2\) is equal to
- A 10
- B -10
- C 26
- D -26
Answer & Solution
Correct Answer
(C) 26
Step-by-step Solution
Detailed explanation
Given, \(x^3-3 x^2-4 x+12=0\)…
See the Complete Solution
Get step-by-step explanations for this and 2.5 Lakh+ more JEE, NEET & CET questions.
- Unlock all solutions
- Practice the full chapter
- Track accuracy across PYQs
4.8 rated on Google Play · 14,000+ reviews
More questions from Maths
- TS EAMCET 2021 Medium
- If the length of the chord \(2 x+3 y+\mathrm{k}=0\) of the circle \(x^2+y^2-2 x+4 y-11=0\) is \(2 \sqrt{3}\), then the sum of all possible values of \(k\) isTS EAMCET 2025 Medium
- If denotes the area of triangle , thenTS EAMCET 2022 Easy
- Let and . ThenTS EAMCET 2022 Easy
- In any triangle \(\mathrm{ABC}, r^2 \cot \frac{\mathrm{A}}{2} \cot \frac{\mathrm{B}}{2} \cot \frac{\mathrm{C}}{2}=\)TS EAMCET 2022 Easy
- If \(x, y, z\) are all positive and are the \(p\) th, \(q\) th and \(r\) th terms of a geometric progression respectively, then the value of the determinant
\(\left|\begin{array}{lll}
\log x & p & 1 \\
\log y & q & 1 \\
\log z & r & 1
\end{array}\right| \text { equals }\)TS EAMCET 2009 Medium
More PYQs from TS EAMCET
- To remove the second term of the equation \(x^4-8 x^3+x^2-x+3=0\), diminish the roots of the equation byTS EAMCET 2002 Medium
- In an acute angled triangle, \(\cot B \cot C+\cot A \cot C+\cot A \cot B\) is equal toTS EAMCET 2012 Easy
- The emf (in V) of a Daniell cell containing \(0.1 \mathrm{MZnSO}_4\) and \(0.01 \mathrm{M} \mathrm{CuSO}_4\) solutions at their respective electrodes is \(\left(E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=+0.34 \mathrm{~V} ; E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}=-0.76 \mathrm{~V}\right)\)TS EAMCET 2012 Easy
- Which pair of oxyacids of phosphorus contains ' \(\mathrm{P}-\mathrm{H}\) ' bonds?TS EAMCET 2009 Medium
- For the gaseous reaction, \(\mathrm{N}_2 \mathrm{O}_5 \rightarrow 2 \mathrm{NO}_2+\frac{1}{2} \mathrm{O}_2\) the rate can be expressed as
\( -\frac{\mathrm{d}\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\mathrm{dt}}=\mathrm{K}_1\left[\mathrm{~N}_2 \mathrm{O}_5\right] \ +\frac{\mathrm{d}\left[\mathrm{NO}_2\right]}{\mathrm{dt}}=\mathrm{K}_2\left[\mathrm{~N}_2 \mathrm{O}_5\right] \ +\frac{\mathrm{d}\left[\mathrm{O}_2\right]}{\mathrm{dt}}=\mathrm{K}_3\left[\mathrm{~N}_2 \mathrm{O}_5\right]\) The correct relation between \(\mathrm{K}_1, \mathrm{~K}_2\) and \(\mathrm{K}_3\) isTS EAMCET 2024 Easy - If the circles \(x^2+y^2-2 \lambda x-2 y-7=0\) and \(3\left(x^2+y^2\right)-8 x+29 y=0\) are orthogonal, then \(\lambda\) is equal toTS EAMCET 2016 Easy