Irrational numbers and real numbers

(\sqrt{8}=2\sqrt{2}), so the sum is (\sqrt{2}-2\sqrt{2}=-\sqrt{2}). Simplifying radicals first reduces mistakes.

Using the formula, (x=\frac{2\sqrt{3}\pm\sqrt{12+4}}{2}=\sqrt{3}\pm2). Simplify (\sqrt{16}=4) carefully.

The product is (ab=\sqrt{2}\cdot\sqrt{18}=\sqrt{36}=6). In radical multiplication, simplify the product inside the root first.

By the formula, (x=\frac{-4\pm\sqrt{16-8}}{2}=-2\pm\sqrt{2}). Pay attention to the negative sign and denominator (2).

((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta=3^2-4(-2)=17). This method gives the answer without finding the zeroes.

The product is (5), so the other zero is (\frac{5}{\sqrt{5}}=\sqrt{5}). The sum is (2\sqrt{5}=2k), hence (k=\sqrt{5}).

For real zeroes, the discriminant must be positive, and for irrational zeroes it must not be a perfect square. This is the key check for quadratics with rational coefficients.

(\alpha+\beta=2) and (\alpha\beta=-11), so (\alpha^2+\beta^2+\alpha\beta=(\alpha+\beta)^2-\alpha\beta=4+11=15). Sum and product are enough for symmetric expressions.

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 \((x-3)(x-\sqrt{2})\).

To find the zeroes, set p(x)=0: x²−16=0. This is a difference of two squares, so (x−4)(x+4)=0. Therefore x=4 or x=−4. Both roots are integers, and integers are rational numbers; they are also real because they lie on the real number line. Hence both zeroes are rational real numbers, as stated in option A. The presence of a square does not automatically make a root irrational: √16 simplifies to 4. Option B incorrectly leaves √16 unsimplified, option C would require a negative discriminant, and option D wrongly assigns different types to the two roots.

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.

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

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

For a monic quadratic x²+px−3 with zeroes α and β, Vieta’s relations give α+β=−p and αβ=−3. Add the two given zeroes: (2+√7)+(2−√7)=4, because the irrational terms cancel. Therefore −p=4, so p=−4. The product independently checks the information: (2+√7)(2−√7)=2²−(√7)²=4−7=−3, which agrees with the constant term. Hence option A is correct. Option B wrongly treats the sum as p rather than −p. Options C and D confuse p with the constant term −3 or mishandle its sign. Both root relations confirm the same answer.

For equal zeroes, (D=0), so (4-4n=0) and (n=1). Then the zero is (1), which is not irrational.

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

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.

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.

The sum is (\sqrt{2}+\sqrt{8}=\sqrt{2}+2\sqrt{2}=3\sqrt{2}). Simplify radicals before giving the final answer.

For (p(x)), (D=121-96=25), a perfect square. For (q(x)), (D=121-92=29), positive and not a perfect square.