Irrational numbers and real numbers

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

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.

Direct answer: option A, no such real \(n\) exists. Equal zeroes require discriminant \(D=0\). Here \(D=(-2)^2-4(1)(n)=4-4n\), so \(4-4n=0\) gives \(n=1\). The polynomial then becomes \(x^2-2x+1=(x-1)^2\), whose equal zeroes are both 1. But 1 is rational, not irrational. Therefore no real value of \(n\) can make the zeroes both equal and irrational. B gives equal zeroes but they are rational. C gives a negative constant and does not satisfy the required condition. D gives \(D=4-8=-4\), so the zeroes are not real. Thus A is correct. Memory cue: equal roots mean \(D=0\), then check the nature of that root.

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 direct answer is option A: the sum is \(3\sqrt{2}\). For \(x^2-Sx+P\), the coefficient of x is the negative of the sum of the zeroes, so the sum here is \(\sqrt{2}+\sqrt{8}\). Simplify \(\sqrt{8}=\sqrt{4\cdot2}=2\sqrt{2}\). Then add like surds: \(\sqrt{2}+2\sqrt{2}=3\sqrt{2}\). Option A matches. Option B, \(2\sqrt{2}\), forgets the original \(\sqrt{2}\). Option C, \(4\sqrt{2}\), adds incorrectly. Option D, \(\sqrt{10}\), treats the sum of radicals as though their radicands could simply be multiplied; that is not a valid rule. The constant 4 is used for the product of zeroes, not their sum. Memory cue: first simplify every radical, then combine like radicals.

The direct answer is option A. For a quadratic \(ax^2+bx+c\), the discriminant is \(D=b^2-4ac\). It tells us the nature of the roots. For \(p(x)=x^2-11x+24\), \(D=(-11)^2-4(1)(24)=121-96=25\). Since 25 is a positive perfect square, the roots are real, rational, and distinct. For \(q(x)=x^2-11x+23\), \(D=121-4(1)(23)=121-92=29\). Since 29 is positive but not a perfect square, its roots are real and irrational. Thus option A is correct. Option B is wrong because q has irrational roots. Option C is wrong because both discriminants are positive, so the roots are not non-real. Option D reverses the two conclusions. Exam cue: positive square discriminant means rational roots; positive nonsquare discriminant means irrational real roots.

For a quadratic \(ax^2+bx+c\) with roots \(\alpha,\beta\), the difference is \(|\alpha-\beta|=\dfrac{\sqrt{D}}{|a|}\) where \(D=b^2-4ac\). Here \(a=1,\;b=-6\sqrt{2},\;c=17\), so \(D=(6\sqrt{2})^2-4\cdot1\cdot17=72-68=4\). Hence \(|\alpha-\beta|=\sqrt{4}=2\). The common distractor \(4\) is the discriminant itself, not the root difference. Exam tip: use \(|\alpha-\beta|=\sqrt{(\alpha+\beta)^2-4\alpha\beta}\) or \(|\alpha-\beta|=\dfrac{\sqrt{D}}{|a|}\) to get the answer quickly.

For a monic quadratic \(x^2-Sx+P\), the sum of the zeroes is \(S\), and their product is \(P\). The two zeroes are \(2\sqrt3+1\) and \(2\sqrt3-1\). Adding them cancels the 1 and \(-1\): \((2\sqrt3+1)+(2\sqrt3-1)=4\sqrt3\). Multiplying them uses the difference-of-squares identity: \((2\sqrt3)^2-1^2=12-1=11\).

Hence \(S=4\sqrt3\) and \(P=11\), which is exactly option A. The signs in the polynomial are important: in \(x^2-Sx+P\), the coefficient notation already makes the sum equal to \(S\), rather than \(-S\). The supplied answer is therefore correct and follows directly from the standard relations between roots and coefficients.

With rational coefficients, irrational parts occur in conjugate pairs. Only (6+\sqrt{5}) and (6-\sqrt{5}) have both rational sum and rational product.

((1+\sqrt{3})^2-2(1+\sqrt{3})-2=1+2\sqrt{3}+3-2-2\sqrt{3}-2=0). Do not forget the middle term while expanding the square.

Since (p(1+\sqrt{3})=0), (1+\sqrt{3}) is a zero. To prove a number is a zero, show that the polynomial value is (0).

The sum of zeroes is (2), so the other zero is (2-(1+\sqrt{3})=1-\sqrt{3}). With rational coefficients, the conjugate also appears.

With roots \(a+\sqrt{b}\) and \(a-\sqrt{b}\), the sum of roots is \(2a\) and the product is \(a^{2}-b\). The monic quadratic is \(x^{2}-2ax+(a^{2}-b)\). Its discriminant is \(\Delta=(2a)^{2}-4\cdot1\cdot(a^{2}-b)=4a^{2}-4a^{2}+4b=4b\), so \(4b\) is correct. Closest distractor D equals \(4(a^{2}-b)=4a^{2}-4b\), which is generally different from \(4b\); option C ignores the subtraction in the product and B would only hold if \(b=0\). Exam tip: use sum and product of roots to form the quadratic quickly, then compute \(\Delta\) from its coefficients.

Here (\alpha+\beta=8) and (\alpha\beta=1), so (\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}=\frac{64-2}{1}=62). In such questions, first find the sum and product.

With rational coefficients, (a+\sqrt{b}) is accompanied by (a-\sqrt{b}). In exams identify conjugate zeroes quickly.

Core idea: conjugate pairs use the difference-of-squares identity \((a+b)(a-b)=a^2-b^2\). Here \(a=5\) and \(b=\sqrt{3}\), so the product is \((5+\sqrt{3})(5-\sqrt{3})=5^2-(\sqrt{3})^2=25-3=22\). Option D (\(25+\sqrt{3}\)) is incorrect because it is a sum not the result of difference of squares; B and C arise from simple arithmetic mistakes. Exam tip: on seeing conjugate zeros, apply \((a+b)(a-b)\) immediately to avoid extra steps and errors.

The sum is (6) and the product is (7), so the polynomial is (x^2-6x+7). In exams use (x^2-(\text{sum})x+\text{product}).

Expand the square: x^2 = (\sqrt{6}+\sqrt{2})^2 = 6 + 2 + 2\sqrt{12} = 8 + 2\times 2\sqrt{3} = 8 + 4\sqrt{3}. Option B (8+2\sqrt{3}) is a common error caused by not fully simplifying \sqrt{12} (since \sqrt{12}=2\sqrt{3}, the cross-term becomes 4\sqrt{3}, not 2\sqrt{3}). Exam tip: apply (a+b)^2 = a^2+2ab+b^2 and simplify radicals fully before choosing the answer.

The denominator contains (7), so the decimal will not terminate and being rational it will recur. In exams decide after checking the denominator in lowest form.

In (\frac{63}{2^5\cdot5^2\cdot7}), after cancelling (63) and (7), only (2) and (5) remain in the denominator. In exams reduce the fraction first.

These are conjugate terms. \(ab=(\sqrt{11}+\sqrt{5})(\sqrt{11}-\sqrt{5})=(\sqrt{11})^2-(\sqrt{5})^2=11-5=6\). A common wrong choice is \(\sqrt{55}\), which equals \(\sqrt{11}\cdot\sqrt{5}\) but is not the product of the conjugates. Option 16 arises from adding 11 and 5 (not correct for multiplication), and \(2\sqrt{55}\) is an incorrect doubled product. Exam tip: recognize conjugates and apply the difference-of-squares identity \(a^2-b^2\) to remove radicals quickly.