Irrational numbers and real numbers

(\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.

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.

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}).

A zero of a polynomial is a value of x for which the polynomial becomes zero. Set 5x²−5=0. Factoring gives 5(x²−1)=0, so x²−1=0 and therefore (x−1)(x+1)=0. Hence x=1 or x=−1. Both numbers are integers, and every integer is rational because it can be written as a quotient of two integers, such as 1=1/1 and −1=−1/1. The common factor 5 does not produce √5; it is removed by division. Thus option A is correct. Option C confuses coefficients with roots, option B results from an incorrect rearrangement, and option D is false because two real roots exist.

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.

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.

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.

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

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

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

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.

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.

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.

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

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.

For a quadratic, the discriminant is \(D=b^2-4ac\). Here \(a=1,\;b=-2,\;c=5\), so \(D=(-2)^2-4\cdot1\cdot5=4-20=-16\), which is negative. When \(D<0\) the quadratic has no real roots; instead it has a pair of complex conjugate roots. The closest distractor \(D>0\) and not a perfect square would imply two distinct real (typically irrational) roots, so it does not apply here. Exam tip: compute \(D\) quickly — sign of \(D\) decides real vs. repeated vs. complex roots.

(\sqrt{12}=\sqrt{4\cdot3}=2\sqrt{3}). While simplifying zeroes, take square factors outside the radical.

Use Vieta’s relations for a quadratic ax²+bx+c. If its zeroes are α and β, then α+β=−b/a and αβ=c/a. In p(x)=x²−2√6x+5, the coefficients are a=1, b=−2√6, and c=5. Hence the sum is −(−2√6)/1=2√6, while the product is 5/1=5. Thus option A gives both values in the requested order. Option B interchanges the sum and product. Option C incorrectly changes the sign of the product even though the constant term is positive. Option D forgets the negative sign in the formula for the sum. Individual roots are unnecessary because the coefficient relations answer the question directly.

The polynomial equals ((x-a)^2-b), so (x=a\pm\sqrt{b}). When (b) is not a perfect square, (\sqrt{b}) is irrational.

For (p(x)), (D=81-56=25), a perfect square, so the zeroes are rational. For (q(x)), (D=81-60=21), positive but not a perfect square, so the zeroes are irrational real.