Irrational numbers and real numbers

The sum (\sqrt{2}+\sqrt{3}) and product (\sqrt{6}) match the option (\sqrt{2}), (\sqrt{3}). Hence those are the zeroes.

The sum is (2-\sqrt{5}), so the coefficient of (x) is (-(2-\sqrt{5})=\sqrt{5}-2). The product (-2\sqrt{5}) also matches.

Both zeroes are (\sqrt{3}), so the sum is (2\sqrt{3}). In (x^2+kx+3), the sum is (-k), hence (k=-2\sqrt{3}).

For rational zeros of a quadratic with integer coefficients, the discriminant \(D=b^2-4ac\) must be a perfect square. Here \(a=1, b=-4, c=r\). For \(r=3\): \(D=(-4)^2-4\cdot1\cdot3=16-12=4\), which is a perfect square (\(\sqrt{D}=2\)). Hence the zeros are \((4\pm2)/2\), i.e. 3 and 1, both rational. Distractor B is incorrect because the discriminant is not non-square; C and D are wrong because \(D<0\) (complex roots) and \(D=0\) (equal roots) do not hold here — actually \(D=4>0\). Exam tip: compute \(D\) first, check if it is a perfect square, then compute the roots to confirm.

For \((x^2-6x+7)\) the discriminant is \(D=b^2-4ac=36-28=8\). Since \(D>0\) and not a perfect square, the roots are \(3\pm\sqrt{2}\) — irrational and conjugate. The coefficients (1, −6, 7) are rational. Why other choices are wrong: \((x^2-\sqrt{2}x+1)\) has an irrational coefficient (so coefficients are not all rational); \((x^2-4x+4)\) has \(D=0\) giving a repeated rational root 2; \((x^2+1)\) has \(D<0\) giving purely imaginary roots. Exam tip: check coefficient rationality first, then compute the discriminant — irrational conjugate real roots occur when discriminant is positive but not a perfect square.

(\alpha+\beta=4) and (\alpha\beta=4-7=-3). Thus (\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=16+6=22).

(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}). Here the sum is (10) and product is (25-6=19), so the answer is (\frac{10}{19}).

Direct answer: option A, \(\frac{10}{19}\). Let the zeroes be \(\alpha,eta\). For \(x^2-10x+19\), the sum is \(\alpha+eta=10\), and the product is \(\alpha\beta=19\). Now use the fraction rule: \(\frac1\alpha+\frac1\beta=\frac{\alpha+\beta}{\alpha\beta}=\frac{10}{19}\). The denominator is nonzero because the product is 19. A is correct. B reverses numerator and denominator. C is only the sum, not the reciprocal sum. D is only the product. The key step is to combine fractions before substituting. Memory cue: reciprocal sum equals sum divided by product.

Direct answer: option A, \(x^2-12x+16\). With rational coefficients, an irrational zero containing \(\sqrt5\) brings its conjugate \(6+2\sqrt5\) as the other zero. Let the zeros be \(\alpha=6-2\sqrt5\) and \(\beta=6+2\sqrt5\). Their sum is 12, and their product is \(36-(2\sqrt5)^2=36-20=16\). A monic quadratic with zeros \(\alpha,\beta\) is \(x^2-(\alpha+\beta)x+\alpha\beta=x^2-12x+16\). Thus A works. B has the wrong sum and product. C has the wrong signs and does not have these positive zeros. D has the correct sum but product 20 instead of 16. The conjugate-root rule is essential for rational-coefficient polynomials. Memory cue: sum gives the negative middle coefficient; product gives the constant term.

From (x^2-2=0), (x=\pm\sqrt{2}), which are irrational real numbers. In exams, check both real nature and rationality of roots.

The sum is (3+\sqrt{5}+3-\sqrt{5}=6), which is rational. Conjugate irrational numbers often have a rational sum.

The discriminant is (D=36-16=20), so the zeroes are (3\pm\sqrt{5}). If (D) is not a perfect square, real zeroes can be irrational.

With rational coefficients, the conjugate of an irrational zero is also a zero. So (2-\sqrt{3}) will be the other zero.

The sum of zeroes is (0) and product is (-7), so the polynomial is (x^2-7). Use (x^2-\text{sum}x+\text{product}) to form a polynomial from zeroes.

The product is (\frac{c}{a}=\frac{1}{1}=1), which is rational. Zeroes may be irrational but their product can be rational.

Sum of zeros = (5+\sqrt{2})+(5-\sqrt{2}) = 10 and product = (5+\sqrt{2})(5-\sqrt{2}) = 5^2-(\sqrt{2})^2 = 25-2 = 23. For a monic quadratic with zeros α and β the polynomial is x^2 - (α+β)x + αβ. Hence the polynomial is x^2 - 10x + 23. The closest distractor C (x^2 - 10x + 27) only miscalculates the product; B and D have the wrong sign for the sum-term. Exam tip: conjugate irrational zeros give rational coefficients—use sum and product formulas directly to form the polynomial.

By the formula, (x=\frac{4\pm\sqrt{16+8}}{4}=1\pm\frac{\sqrt{6}}{2}). Divide the whole expression carefully while simplifying.

For rational coefficients, the conjugate (a-\sqrt{b}) accompanies (a+\sqrt{b}). Hence the first pair is correct.

Since (p(x)=(x-\sqrt{3})^2), both zeroes are (\sqrt{3}). Recognize perfect-square form for equal zeroes.

(\sqrt{5}+(-\sqrt{5})=0) and (\sqrt{5}\cdot(-\sqrt{5})=-5). Opposite irrationals can have zero sum.

Roots are \\(1+\sqrt{10}\\) and \\(1-\sqrt{10}\\). Their sum is \\((1+\sqrt{10})+(1-\sqrt{10})=2\\) and product is \\((1+\sqrt{10})(1-\sqrt{10})=1-10=-9\\). The monic quadratic with these roots is \\( (x-(1+\sqrt{10}))(x-(1-\sqrt{10}))=x^2-(\text{sum})x+\text{product}=x^2-2x-9\\). Option B is a common distractor because it has the wrong sign for the linear term (would correspond to sum \\(-2\\)). Exam tip: form the polynomial as \\( (x-r_1)(x-r_2)\\) or use sum and product relations \\(r_1+r_2=-\tfrac{b}{a},\;r_1r_2=\tfrac{c}{a}\\).

Since (3x^2-12x+6=3(x^2-4x+2)), the zeroes are (2\pm\sqrt{2}). Removing a common factor first makes calculation easier.

The sum is (-\frac{b}{a}=-4) and (D=16-4=12), not a perfect square. Hence both zeroes are irrational real.

The product is (\frac{(1+\sqrt{3})(1-\sqrt{3})}{4}=\frac{1-3}{4}=-\frac{1}{2}). Use (a^2-b) for conjugate products.

Compute the discriminant: \(D=b^2-4ac=4-4k=4(1-k)\). For real roots we need \(D>0\) (so \(k<1\)). For the roots to be irrational (given rational coefficients), \(D\) must be positive but not a perfect square. For \(k=-1\), \(D=8\), which is positive and not a perfect square, so the roots are real and irrational. Why other choices fail: \(k=0\) gives \(D=4\) (a perfect square) so roots are rational; \(k=1\) gives \(D=0\) (equal rational roots); \(k=2\) gives \(D=-4\) (complex roots). Exam tip: first check sign of \(D\), then check whether \(D\) is a perfect square to decide rational vs irrational roots.