Irrational numbers and real numbers

For a monic quadratic the relation is \(x^2-Sx+P\), where \(S\) is the sum of the roots and \(P\) their product. With \(S=5\sqrt{2}\) and \(P=12\) the polynomial is \(x^2-5\sqrt{2}x+12\). Option B has the wrong sign on the linear term; for \(x^2+5\sqrt{2}x+12\) the sum of roots would be \(-5\sqrt{2}\), not \(+5\sqrt{2}\). Options C and D do not match the given sum and product either. Exam tip: Use \(x^2-(\text{sum})x+(\text{product})\) for monic quadratics and double-check signs of coefficients.

For a quadratic \(ax^2+bx+c\), the product of the zeroes equals \(\frac{c}{a}\). Here \(a=1\) and \(c=\sqrt{2}\), so the product is \(\frac{\sqrt{2}}{1}=\sqrt{2}\). Option B (\(-\sqrt{2}\)) is wrong because the signs of \(a\) and \(c\) are both positive, giving a positive product. Exam tip: always identify \(a,b,c\) first and use \(\frac{c}{a}\) for the product and \(-\frac{b}{a}\) for the sum of roots.

The sum is (-\frac{b}{a}=\frac{4}{\sqrt{2}}=2\sqrt{2}). Rationalising the denominator simplifies the answer.

For a quadratic ax^2+bx+c the product of zeroes is \(\alpha\beta=\frac{c}{a}\). Here a=\(\sqrt{3}\) and c=\(2\sqrt{3}\), so \(\alpha\beta=\frac{2\sqrt{3}}{\sqrt{3}}=2\). Option B (\(\sqrt{3}\)) is incorrect — it results from mishandling the radical cancellation or using wrong coefficients. Exam tip: Use \(\frac{c}{a}\) directly and cancel common radical factors carefully to avoid mistakes.

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}).

The sum of zeroes is (10) and the product is (19). Hence (\frac{1}{\alpha}+\frac{1}{\beta}=\frac{10}{19}).

The other zero is (6+2\sqrt{5}). The sum is (12) and product is (36-20=16), so the polynomial is (x^2-12x+16).

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.