Irrational numbers and real numbers

Correct answer: A. (3, \sqrt{2})

Explanation: For a quadratic \(ax^2+bx+c\), the sum of roots is \(-b/a\) and the product is \(c/a\). Here \(a=1,\; b=-(3+\sqrt{2}),\; c=3\sqrt{2}\). So sum = \(3+\sqrt{2}\) and product = \(3\sqrt{2}\). These match the pair 3 and \(\sqrt{2}\), hence the zeros are 3 and \(\sqrt{2}\). The closest distractor \((3+\sqrt{2},0)\) is wrong because its product would be 0, not \(3\sqrt{2}\). Exam tip: verify roots quickly by checking sum and product or by factoring to \((x-3)(x-\sqrt{2})\).

Correct answer: A. \(x^2-2\)

Explanation: For \(x^2-2=0\), the zeroes are \(\pm\sqrt{2}\), and \(\sqrt{2}\) is irrational; hence both are irrational real numbers. In contrast, \(x^2-4\) has rational zeroes \(\pm2\). Exam tip: check whether the discriminant is a positive non-perfect square.

Correct answer: A. (4)

Explanation: (D=(2\sqrt{7})^2-4\cdot1\cdot6=28-24=4). Even with an irrational coefficient, the discriminant can be rational.

Correct answer: A. (-\sqrt{7}+1) and (-\sqrt{7}-1)

Explanation: Using the formula, (x=\frac{-2\sqrt{7}\pm2}{2}=-\sqrt{7}\pm1). Simplifying the discriminant first gives a clean answer.

Correct answer: A. No such real (n) exists

Explanation: Direct answer: option A, no such real \(n\) exists. Equal zeroes require discriminant \(D=0\). Here \(D=(-2)^2-4(1)(n)=4-4n\), so \(4-4n=0\) gives \(n=1\). The polynomial then becomes \(x^2-2x+1=(x-1)^2\), whose equal zeroes are both 1. But 1 is rational, not irrational. Therefore no real value of \(n\) can make the zeroes both equal and irrational. B gives equal zeroes but they are rational. C gives a negative constant and does not satisfy the required condition. D gives \(D=4-8=-4\), so the zeroes are not real. Thus A is correct. Memory cue: equal roots mean \(D=0\), then check the nature of that root.

Correct answer: A. (\frac{3-\sqrt{5}}{2})

Explanation: With rational coefficients, the conjugate of the irrational part is also a zero. Hence (\frac{3-\sqrt{5}}{2}) is the other zero.

Correct answer: A. \\(x^2-3x+1\\)

Explanation: Let the roots be \\(
\\alpha=\\frac{3+\\sqrt{5}}{2},\\quad \\\beta=\\frac{3-\\sqrt{5}}{2}
\\).\n\nSum: \\(
\\alpha+\\beta=\\frac{3+\\sqrt{5}}{2}+\\frac{3-\\sqrt{5}}{2}=3
\\).\nProduct: \\(
\\alpha\\beta=\\frac{(3+\\sqrt{5})(3-\\sqrt{5})}{4}=\\frac{9-5}{4}=1
\\).\nA monic quadratic with these roots is \\(
x^2-(\text{sum})x+(\text{product})=x^2-3x+1
\\).\nThe closest distractor \\(x^2-3x-1\\) only has the constant term sign wrong, so it is incorrect. Exam tip: use the relation \\(x^2-(r_1+r_2)x+r_1r_2\\) immediately when roots are given.

Correct answer: A. Two distinct real irrational

Explanation: Use the discriminant \(D=b^2-4ac\). Here \(a=1,\;b=-2\sqrt{2},\;c=1\), so \(D=( -2\sqrt{2})^2-4\cdot1\cdot1=8-4=4>0\). Thus the roots are real and distinct. Applying the quadratic formula gives roots \(\frac{-b\pm\sqrt{D}}{2a}=\frac{2\sqrt{2}\pm2}{2}=\sqrt{2}\pm1\). Since \(\sqrt{2}\) is irrational, \(\sqrt{2}\pm1\) are also irrational. Therefore the zeros are two distinct real irrational numbers. The closest distractor (two equal roots) is incorrect because \(D\neq0\). Exam tip: check \(D\) first to determine reality and multiplicity of roots, then inspect any irrational parts (like \(\sqrt{2}\)) to decide rationality.

Correct answer: A. (3\sqrt{2})

Explanation: The direct answer is option A: the sum is \(3\sqrt{2}\). For \(x^2-Sx+P\), the coefficient of x is the negative of the sum of the zeroes, so the sum here is \(\sqrt{2}+\sqrt{8}\). Simplify \(\sqrt{8}=\sqrt{4\cdot2}=2\sqrt{2}\). Then add like surds: \(\sqrt{2}+2\sqrt{2}=3\sqrt{2}\). Option A matches. Option B, \(2\sqrt{2}\), forgets the original \(\sqrt{2}\). Option C, \(4\sqrt{2}\), adds incorrectly. Option D, \(\sqrt{10}\), treats the sum of radicals as though their radicands could simply be multiplied; that is not a valid rule. The constant 4 is used for the product of zeroes, not their sum. Memory cue: first simplify every radical, then combine like radicals.

Correct answer: A. (p(x)) has rational zeroes and (q(x)) has irrational real zeroes

Explanation: The direct answer is option A. For a quadratic \(ax^2+bx+c\), the discriminant is \(D=b^2-4ac\). It tells us the nature of the roots. For \(p(x)=x^2-11x+24\), \(D=(-11)^2-4(1)(24)=121-96=25\). Since 25 is a positive perfect square, the roots are real, rational, and distinct. For \(q(x)=x^2-11x+23\), \(D=121-4(1)(23)=121-92=29\). Since 29 is positive but not a perfect square, its roots are real and irrational. Thus option A is correct. Option B is wrong because q has irrational roots. Option C is wrong because both discriminants are positive, so the roots are not non-real. Option D reverses the two conclusions. Exam cue: positive square discriminant means rational roots; positive nonsquare discriminant means irrational real roots.

Correct answer: A. \(2\)

Explanation: For a quadratic \(ax^2+bx+c\) with roots \(\alpha,\beta\), the difference is \(|\alpha-\beta|=\dfrac{\sqrt{D}}{|a|}\) where \(D=b^2-4ac\). Here \(a=1,\;b=-6\sqrt{2},\;c=17\), so \(D=(6\sqrt{2})^2-4\cdot1\cdot17=72-68=4\). Hence \(|\alpha-\beta|=\sqrt{4}=2\). The common distractor \(4\) is the discriminant itself, not the root difference. Exam tip: use \(|\alpha-\beta|=\sqrt{(\alpha+\beta)^2-4\alpha\beta}\) or \(|\alpha-\beta|=\dfrac{\sqrt{D}}{|a|}\) to get the answer quickly.

Correct answer: A. (S=4\sqrt{3}), (P=11)

Explanation: For a monic quadratic \(x^2-Sx+P\), the sum of the zeroes is \(S\), and their product is \(P\). The two zeroes are \(2\sqrt3+1\) and \(2\sqrt3-1\). Adding them cancels the 1 and \(-1\): \((2\sqrt3+1)+(2\sqrt3-1)=4\sqrt3\). Multiplying them uses the difference-of-squares identity: \((2\sqrt3)^2-1^2=12-1=11\).

Hence \(S=4\sqrt3\) and \(P=11\), which is exactly option A. The signs in the polynomial are important: in \(x^2-Sx+P\), the coefficient notation already makes the sum equal to \(S\), rather than \(-S\). The supplied answer is therefore correct and follows directly from the standard relations between roots and coefficients.

Correct answer: A. Zeroes (6+\sqrt{5}) and (6-\sqrt{5})

Explanation: With rational coefficients, irrational parts occur in conjugate pairs. Only (6+\sqrt{5}) and (6-\sqrt{5}) have both rational sum and rational product.

Correct answer: A. (0)

Explanation: ((1+\sqrt{3})^2-2(1+\sqrt{3})-2=1+2\sqrt{3}+3-2-2\sqrt{3}-2=0). Do not forget the middle term while expanding the square.

Correct answer: A. It is a zero of (p(x))

Explanation: Since (p(1+\sqrt{3})=0), (1+\sqrt{3}) is a zero. To prove a number is a zero, show that the polynomial value is (0).

Correct answer: A. (1-\sqrt{3})

Explanation: The sum of zeroes is (2), so the other zero is (2-(1+\sqrt{3})=1-\sqrt{3}). With rational coefficients, the conjugate also appears.

Correct answer: A. \(4b\)

Explanation: With roots \(a+\sqrt{b}\) and \(a-\sqrt{b}\), the sum of roots is \(2a\) and the product is \(a^{2}-b\). The monic quadratic is \(x^{2}-2ax+(a^{2}-b)\). Its discriminant is \(\Delta=(2a)^{2}-4\cdot1\cdot(a^{2}-b)=4a^{2}-4a^{2}+4b=4b\), so \(4b\) is correct. Closest distractor D equals \(4(a^{2}-b)=4a^{2}-4b\), which is generally different from \(4b\); option C ignores the subtraction in the product and B would only hold if \(b=0\). Exam tip: use sum and product of roots to form the quadratic quickly, then compute \(\Delta\) from its coefficients.

Correct answer: A. (62)

Explanation: Here (\alpha+\beta=8) and (\alpha\beta=1), so (\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}=\frac{64-2}{1}=62). In such questions, first find the sum and product.

Correct answer: A. (2-\sqrt{7})

Explanation: With rational coefficients, (a+\sqrt{b}) is accompanied by (a-\sqrt{b}). In exams identify conjugate zeroes quickly.

Correct answer: A. 22

Explanation: Core idea: conjugate pairs use the difference-of-squares identity \((a+b)(a-b)=a^2-b^2\). Here \(a=5\) and \(b=\sqrt{3}\), so the product is \((5+\sqrt{3})(5-\sqrt{3})=5^2-(\sqrt{3})^2=25-3=22\). Option D (\(25+\sqrt{3}\)) is incorrect because it is a sum not the result of difference of squares; B and C arise from simple arithmetic mistakes. Exam tip: on seeing conjugate zeros, apply \((a+b)(a-b)\) immediately to avoid extra steps and errors.

Correct answer: A. (x^2-6x+7)

Explanation: The sum is (6) and the product is (7), so the polynomial is (x^2-6x+7). In exams use (x^2-(\text{sum})x+\text{product}).

Correct answer: A. 8+4\sqrt{3}

Explanation: Expand the square: x^2 = (\sqrt{6}+\sqrt{2})^2 = 6 + 2 + 2\sqrt{12} = 8 + 2\times 2\sqrt{3} = 8 + 4\sqrt{3}. Option B (8+2\sqrt{3}) is a common error caused by not fully simplifying \sqrt{12} (since \sqrt{12}=2\sqrt{3}, the cross-term becomes 4\sqrt{3}, not 2\sqrt{3}). Exam tip: apply (a+b)^2 = a^2+2ab+b^2 and simplify radicals fully before choosing the answer.

Correct answer: A. Non-terminating recurring

Explanation: The denominator contains (7), so the decimal will not terminate and being rational it will recur. In exams decide after checking the denominator in lowest form.

Correct answer: A. (\frac{63}{2^5\cdot5^2\cdot7})

Explanation: In (\frac{63}{2^5\cdot5^2\cdot7}), after cancelling (63) and (7), only (2) and (5) remain in the denominator. In exams reduce the fraction first.

Correct answer: A. 6

Explanation: These are conjugate terms. \(ab=(\sqrt{11}+\sqrt{5})(\sqrt{11}-\sqrt{5})=(\sqrt{11})^2-(\sqrt{5})^2=11-5=6\). A common wrong choice is \(\sqrt{55}\), which equals \(\sqrt{11}\cdot\sqrt{5}\) but is not the product of the conjugates. Option 16 arises from adding 11 and 5 (not correct for multiplication), and \(2\sqrt{55}\) is an incorrect doubled product. Exam tip: recognize conjugates and apply the difference-of-squares identity \(a^2-b^2\) to remove radicals quickly.