Irrational numbers and real numbers

For a quadratic ax²+bx+c with zeroes α and β, Vieta’s relation gives αβ=c/a. In p(x)=x²−2x−3√2, a=1 and c=−3√2. Therefore αβ=(−3√2)/1=−3√2, so option A is correct. The sum follows a different relation, α+β=−b/a; here it equals 2, not −3√2. The discriminant is b²−4ac, which becomes (−2)²−4(1)(−3√2)=4+12√2, so option D is also incorrect. Finally, an irrational constant term by itself does not prove that both zeroes are rational or irrational; more information would be needed. The key point is that the monic constant term equals the product of the zeroes.

For a quadratic ax²+bx+c with zeroes α and β, the governing relations are α+β=−b/a and αβ=c/a. In p(x)=x²−6√2x+17, the coefficients are a=1, b=−6√2 and c=17. Hence α+β=−(−6√2)/1=6√2, while αβ=17/1=17. Therefore option A is correct. Option B keeps the wrong sign for the sum. Option C swaps the coefficient and constant-term roles, and option D gives the product the wrong sign. The presence of an irrational coefficient does not change Vieta’s relations; they apply to quadratic polynomials regardless of whether the coefficients are rational or irrational.

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.

If (x) were rational then (\sqrt{2}+x) would be irrational. So (x) must be irrational; remember the sum rule for rational and irrational numbers.

The product of a non-zero rational number and an irrational number remains irrational. In exams the condition (a\ne0) is very important.

Here (\sqrt{50}=5\sqrt{2}) and (\sqrt{8}=2\sqrt{2}), so the difference is (3\sqrt{2}), irrational; no listed expression is rational, so this item must be checked carefully.

Since (\sqrt{8}\times\sqrt{18}=\sqrt{144}=12), it is rational. In exams simplify products of radicals first.

(p(\sqrt{2})=2-5\sqrt{2}+6=8-5\sqrt{2}), which is irrational. In exams substitute first and then simplify the radical.

For a quadratic with rational coefficients, if (a+\sqrt{b}) is a zero then (a-\sqrt{b}) is also a zero. The conjugate-root rule is useful in exams.

Compute the sum: \((3+\sqrt{5})+(3-\sqrt{5})=3+3+\sqrt{5}-\sqrt{5}=6\). For conjugate irrational roots the radical parts cancel, leaving the sum of the rational parts. A common wrong choice (option D = 10) results from incorrectly adding or double-counting the radical terms; note that the product of the roots is \((3+\sqrt{5})(3-\sqrt{5})=9-5=4\), which is a different quantity. Exam tip: When roots are conjugates, add the rational parts first — the square-root terms cancel out.

Compute directly using product of conjugates: \((4+\sqrt{7})(4-\sqrt{7})=4^2-(\sqrt{7})^2=16-7=9\). Option B is incorrect — it resembles a sum, not the product. Option C arises from the incorrect step \(4^2+7\) being treated as the product, and D mistakes the product for just \(\sqrt{7}\)-related term. Exam tip: for conjugate zeros use \((a+b)(a-b)=a^2-b^2\); alternatively Vieta's formula gives product = constant term / leading coefficient for quadratics.

The sum is (4) and the product is (4-6=-2), so the polynomial is (x^2-4x-2). Remember the formula (x^2-Sx+P).

The governing idea is direct substitution together with the square-root identity (√a)²=a for every non-negative real number a. Since x=√3, replace x in the expression by √3: x²−3=(√3)²−3. Squaring the complete radical gives (√3)²=3, so the expression becomes 3−3=0. Therefore option B is correct. Option A would be obtained by treating the expression as a quotient of identical quantities, which it is not. Option C gives only x² and forgets the subtraction of 3, while option D adds instead of subtracting. The important step is to square the entire √3, not just manipulate the radicand incorrectly.