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TG EAMCET 2 May 2025 Shift 1 Paper

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Question : 5 of 160

Marks: +1, -0

If

\begin{vmatrix} 9 & 25 & 16 \\ 16 & 36 & 25 \\ 25 & 49 & 36 \end{vmatrix}=K

K, K+1

are the roots of the equation

$x^2-13x+42=0$
$x^2-15x+56=0$
$x^2-19x+90=0$
$x^2-17x+72=0$

Solution:

\;\; \text{ Given, } \; K=\begin{vmatrix} 9 & 25 & 16 \\ 16 & 36 & 25 \\ 25 & 49 & 36 \end{vmatrix}

\; =9 \begin{vmatrix} 36 & 25 \\ 49 & 36 \end{vmatrix} -25 \begin{vmatrix} 16 & 25 \\ 25 & 36 \end{vmatrix} +16 \begin{vmatrix} 16 & 36 \\ 25 & 49 \end{vmatrix}

\; =9(1296-1225)-25(576-625)+16(784-900)

\; =9(71)-25(-49)+16(-116)

\; =639+1225-1856=8

\; \therefore K=8

So, the roots are

K

and

K+1

, i.e., 8 and 9

\; \text{ Sum of roots } \;\; =8+9=17

\; =\; \frac{-b}{a} = \alpha+\beta

And product of roots

=8 \times 9=72

=\; \frac{c}{a} = \alpha \beta

So, the general form of a quadratic equation is

x^2-(\alpha+\beta) x+\alpha \beta=0

\Rightarrow \;\; x^2-17x+72=0

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