Quadratic Equations Questions and Answers

Q: Let \(r\) be a root of the equation \(x^{2} + 2x + 6 = 0\).Then the value of the expression \((r + 2)(r + 3)(r + 4)(r + 5)\) is

The value is \(-126\).Since \(r\) is a root, \(r^{2} + 2r + 6 = 0\), so \(r^{2} + 2r = -6\). Pair the factors so that each product can be reduced with this relation.\[ \begin{aligned} (r+2)(r+3) &= r^{2} + 5r + 6 = 3r \\ (r+4)(r+5) &= r^{2} + 9r + 20 = 7r + 14 \\ \text{product} &= 3r(7r + 14) \\ &= 21(r^{2} + 2r) = 21 \times (-6) = -126 \end{aligned} \]The roots themselves are complex (\(r = -1 \pm i\sqrt{5}\)), so substituting them directly is slow; the reduction avoids that. 126 has the wrong sign, and 51 and \(-51\) come from slips such as reducing \(r^{2}\) to \(-6\) instead of \(-2r - 6\).

Q: The sum of a rational number and its reciprocal is 13/6, find the number

The number is 2/3 or 3/2.Let the number be \(x\). Then:\[ \begin{aligned} x + \tfrac{1}{x} &= \tfrac{13}{6} \\ 6x^{2} - 13x + 6 &= 0 \\ (2x - 3)(3x - 2) &= 0 \end{aligned} \]so \(x = 3/2\) or \(2/3\). Check: \(2/3 + 3/2 = 4/6 + 9/6 = 13/6\). 3/4 + 4/3 = 25/12 and 2/5 + 5/2 = 29/10, so those pairs fail.

If \(6x^2 - 4xy - 10y^2 = 0\) then what is \(x : y\)?

A.\((1 : -1)\) or \((3 : 5)\)
B.\((-1 : 1)\) or \((5 : 3)\)
C.\((-2 : 1)\) or \((5 : 3)\)
D.\((-2 : 1)\) or \((3 : 5)\)

Exam Year(s)

DSSSB PGT - 2023,

Answer: B

Divide the equation by 2 and factorise it as a quadratic in x with y as the parameter.

\[ 3x^2 - 2xy - 5y^2 = 0 \Rightarrow (3x - 5y)(x + y) = 0 \]

The first factor gives \(3x = 5y\), so \(x : y = 5 : 3\). The second factor gives \(x = -y\), so \(x : y = -1 : 1\). So the ratio is \((-1 : 1)\) or \((5 : 3)\). The options with 3 : 5 have the ratio the wrong way round, and the ones with \(-2 : 1\) do not satisfy the equation (for \(x = -2y\) the left side is \(12 + 8 - 10 = 10\) times \(y^2\), which is not zero). The answer is option B.

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Let \(r\) be a root of the equation \(x^{2} + 2x + 6 = 0\).

Then the value of the expression \((r + 2)(r + 3)(r + 4)(r + 5)\) is

A.51
B.−51
C.126
D.−126

Exam Year(s)

GATE CS - 2022,

Answer: D

The value is \(-126\).

Since \(r\) is a root, \(r^{2} + 2r + 6 = 0\), so \(r^{2} + 2r = -6\). Pair the factors so that each product can be reduced with this relation.

\[ \begin{aligned} (r+2)(r+3) &= r^{2} + 5r + 6 = 3r \\ (r+4)(r+5) &= r^{2} + 9r + 20 = 7r + 14 \\ \text{product} &= 3r(7r + 14) \\ &= 21(r^{2} + 2r) = 21 \times (-6) = -126 \end{aligned} \]

The roots themselves are complex (\(r = -1 \pm i\sqrt{5}\)), so substituting them directly is slow; the reduction avoids that. 126 has the wrong sign, and 51 and \(-51\) come from slips such as reducing \(r^{2}\) to \(-6\) instead of \(-2r - 6\).

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The sum of a rational number and its reciprocal is 13/6, find the number

A.2/3 or 3/2
B.3/4 or 4/3
C.2/5 or 5/2
D.None of these

Exam Year(s)

ISRO Assistant - 2023,

Answer: A

The number is 2/3 or 3/2.

Let the number be \(x\). Then:

\[ \begin{aligned} x + \tfrac{1}{x} &= \tfrac{13}{6} \\ 6x^{2} - 13x + 6 &= 0 \\ (2x - 3)(3x - 2) &= 0 \end{aligned} \]

so \(x = 3/2\) or \(2/3\). Check: \(2/3 + 3/2 = 4/6 + 9/6 = 13/6\). 3/4 + 4/3 = 25/12 and 2/5 + 5/2 = 29/10, so those pairs fail.

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Quadratic Equations Questions and Answers

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Quadratic Equations Questions and Answers

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