Irrational numbers and real numbers

Correct answer: A. First are rational, second are irrational real

Explanation: For the first, (D=64-28=36) is a perfect square; for the second, (D=64-40=24) is not. The discriminant quickly tells the type of zeroes.

Correct answer: A. (a^2-b)

Explanation: In a monic polynomial, the constant term is the product of zeroes. Here the product is ((a+\sqrt{b})(a-\sqrt{b})=a^2-b).

Correct answer: C. (x^2-8)

Explanation: Direct answer: option C. Zeroes are found by putting each polynomial equal to zero. For A, \(x^2-9=0\) gives \(x=\pm3\), both rational, so A fails. For B, \(x^2+9=0\) gives \(x^2=-9\), so there are no real zeroes; B fails. For C, \(x^2-8=0\), hence \(x=\pm\sqrt8=\pm2\sqrt2\). These are real because they are square roots of a positive number, and irrational because 8 is not a perfect square. Thus C works. For D, \(x^2-6x+9=(x-3)^2\), so the only zero is 3, rational. Memory cue: positive non-square under a square root gives irrational real roots.

Correct answer: B. At least one coefficient will be irrational

Explanation: The sum (\sqrt{2}+\sqrt{3}) is irrational, so the coefficient of (x) in the monic polynomial is irrational. For rational coefficients, such zeroes must occur as conjugates.

Correct answer: B. (1) and (\sqrt{3})

Explanation: The sum in the given polynomial is (2+\sqrt{3}), while checking options shows a mismatch if done carelessly. This item needs coefficient matching with both sum and product.

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

Explanation: The sum of zeroes is (1+\sqrt{3}) and the product is (\sqrt{3}). The numbers (1) and (\sqrt{3}) satisfy both conditions.

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

Explanation: (p(\sqrt{2})=(\sqrt{2})^2-2\sqrt{2}-2=2-2\sqrt{2}-2=-2\sqrt{2}). When substituting, write ((\sqrt{2})^2=2).

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

Explanation: From (3+a\sqrt{3}+3=0), (a\sqrt{3}=-6), so (a=-2\sqrt{3}). After substituting an irrational value, separate like terms carefully.

Correct answer: A. (x=\pm\sqrt{5})

Explanation: The direct answer is option A: the graph cuts the x-axis at \(x=\pm\sqrt{5}\). A point on the x-axis has y-coordinate zero. Since \(y=p(x)=x^2-5\), put \(y=0\): \(x^2-5=0\), so \(x^2=5\). Taking both square roots gives \(x=\sqrt{5}\) and \(x=-\sqrt{5}\). These are the two zeroes and therefore the two x-intercepts. Option A states both values, so it is correct. Option B, \(x=\pm5\), is wrong because squaring 5 gives 25, not 5. Option C, \(x=0\), is wrong because \(p(0)=-5\), not zero. Option D, \(x=\sqrt{25}=5\), gives only one incorrect value and misses the negative root. Exam cue: set the polynomial equal to zero to find x-axis intersections.

Correct answer: A. p = -8, q = 9

Explanation: The sum of the roots is \((4+\sqrt{7})+(4-\sqrt{7})=8\) and their product is \((4+\sqrt{7})(4-\sqrt{7})=16-7=9\). For the monic quadratic \(x^2+px+q\), sum of roots = \(-p\) and product = \(q\). So \(-p=8\) gives \(p=-8\) and \(q=9\). Hence A is correct. Option B flips the sign of p, while C and D have the wrong product q. Exam tip: use sum = -p and product = q or expand \((x-\alpha)(x-\beta)\) to find coefficients quickly.

Correct answer: A. (2\pm\sqrt{10})

Explanation: By the formula, (x=\frac{4\pm\sqrt{16+24}}{2}=2\pm\sqrt{10}). Remember (\sqrt{40}=2\sqrt{10}) while simplifying (D).

Correct answer: A. (40)

Explanation: (\alpha+\beta=6) and (\alpha\beta=9-11=-2), so (\alpha^2+\beta^2=36-2(-2)=40). The identity (\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta) is useful.

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

Explanation: The sum is (2m) and product is (m^2-3), matching (m+\sqrt{3}) and (m-\sqrt{3}). Even in general form, match sum and product.

Correct answer: A. The other zero will be (-\sqrt{13})

Explanation: For rational coefficients, the conjugate (-\sqrt{13}) of (\sqrt{13}) also appears when the linear coefficient is rational. This follows from (a+\sqrt{b}) and (a-\sqrt{b}).

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

Explanation: The zeroes are (6\pm\sqrt{5}), so the difference is (2\sqrt{5}). The difference of conjugate zeroes is (2) times the radical part.

Correct answer: A. Both irrational real numbers

Explanation: For a quadratic, the discriminant is \(D=b^2-4ac\). Here \(a=1, b=-7, c=5\), so \(D=(-7)^2-4\cdot1\cdot5=49-20=29\). Since \(D>0\) and \(29\) is not a perfect square, the equation has two distinct real irrational roots: \((7\pm\sqrt{29})/2\). Option B is incorrect because a non-perfect-square discriminant does not yield rational roots; option D is incorrect because both roots contain \(\sqrt{29}\) and are therefore both irrational. Exam tip: compute the discriminant first — if positive and not a perfect square, expect two irrational real roots.

Correct answer: A. (x^2-4x+1)

Explanation: In (x^2-4x+1), the sum is (4) and (D=16-4=12), so the zeroes are irrational. A rational sum does not mean rational zeroes.

Correct answer: A. (-\sqrt{5}) twice

Explanation: (p(x)=(x+\sqrt{5})^2), so the zero is (-\sqrt{5}) twice. A perfect-square form gives a repeated zero.

Correct answer: A. (0)

Explanation: The other zero will be (-\sqrt{7}), so the sum is (0) and (a=-0=0). With rational coefficients, take the conjugate zero.

Correct answer: B. One coefficient is irrational

Explanation: The constant term (-\sqrt{2}) is irrational, while the other coefficients are rational. Check coefficient type before applying root rules.

Correct answer: A. √2 and √3

Explanation: Answer: A, √2 and √3. For a monic quadratic, x² − (α + β)x + αβ factors as (x − α)(x − β). Here take α = √2 and β = √3. Their sum is α + β = √2 + √3, exactly the quantity in the middle coefficient. Their product is αβ = √2·√3 = √6, exactly the constant term. Therefore p(x) = (x − √2)(x − √3). A zero is a value of x that makes the polynomial equal to zero, so either factor must be zero: x − √2 = 0 gives x = √2, and x − √3 = 0 gives x = √3. B incorrectly treats the product as one zero and introduces 1 without justification. C mistakes the sum of the zeroes for a zero and also adds 0. D changes both signs; its sum would be negative and would not match the polynomial. Memory cue: in x² − Sx + P, the zeroes have sum S and product P.

Correct answer: A. Zeroes of (p(x)) are rational and zeroes of (q(x)) are irrational real

Explanation: For (p(x)), (D=196-180=16), while for (q(x)), (D=196-160=36), so both are rational. Therefore the listed intended contrast is not valid.

Correct answer: A. Zeroes of (p(x)) are rational and zeroes of (q(x)) are irrational real

Explanation: For (p(x)), (D=16) is a perfect square, and for (q(x)), (D=24) is positive but not a perfect square. Thus the first has rational and the second irrational real zeroes.

Correct answer: A. (-4)

Explanation: (\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{4}{4-5}=-4). Finding sum and product first is easier.

Correct answer: B. Product is (-3), so constant term cannot be (12)

Explanation: For a monic quadratic \(x^2+bx+12\), the product of its zeroes must equal the constant term, because the product is \(\frac{c}{a}\) and here \(a=1\). The proposed zeroes are \(2+\sqrt7\) and \(2-\sqrt7\). Their product is \((2+\sqrt7)(2-\sqrt7)=2^2-(\sqrt7)^2=4-7=-3\). Therefore the constant term should be \(-3\), not 12.

The zeroes are real because both expressions contain the real number \(\sqrt7\), and their sum is \(4\), which is rational. Thus options C and D are false. Option B correctly identifies the inconsistency: the product is \(-3\), so a constant term of 12 cannot be correct. The supplied answer and explanation are mathematically consistent.