Irrational numbers and real numbers

For a quadratic, the discriminant is \(D=b^2-4ac\). Here \(a=1,\;b=-2,\;c=5\), so \(D=(-2)^2-4\cdot1\cdot5=4-20=-16\), which is negative. When \(D<0\) the quadratic has no real roots; instead it has a pair of complex conjugate roots. The closest distractor \(D>0\) and not a perfect square would imply two distinct real (typically irrational) roots, so it does not apply here. Exam tip: compute \(D\) quickly — sign of \(D\) decides real vs. repeated vs. complex roots.

Direct answer: option A, \(\sqrt{12}=2\sqrt3\). A square factor can come outside a square root because \(\sqrt{mn}=\sqrt m\sqrt n\) for non-negative numbers. Write 12 as \(4\times3\): \(\sqrt{12}=\sqrt{4\times3}=\sqrt4\sqrt3=2\sqrt3\). Option A is therefore correct. Option B, \(3\sqrt2\), squares to 18, so it is not \(\sqrt{12}\). Option C, \(6\sqrt2\), is much larger and squares to 72. Option D, \(\sqrt6\), squares to 6, not 12. The best simplification takes the largest perfect-square factor out and leaves no square factor under the radical. Memory cue: search for 4, 9, 16, or another perfect-square factor.

The polynomial equals ((x-a)^2-b), so (x=a\pm\sqrt{b}). When (b) is not a perfect square, (\sqrt{b}) is irrational.

For (p(x)), (D=81-56=25), a perfect square, so the zeroes are rational. For (q(x)), (D=81-60=21), positive but not a perfect square, so the zeroes are irrational real.

A quadratic polynomial with rational coefficients has irrational zeroes in conjugate pairs. Therefore, if 2 + √13 is one zero, the other zero must be 2 − √13. Their sum is (2 + √13) + (2 − √13) = 4, because the irrational parts cancel. For a monic quadratic x² + Bx + C, the sum of the zeroes equals −B, so B = −4. This is confirmed by forming the polynomial: (x − 2 − √13)(x − 2 + √13) = (x − 2)² − 13 = x² − 4x − 9. Hence the coefficient of x is −4, so option A is correct. Option B has the opposite sign, while C and D are not the required rational coefficient.

The discriminant D=b²−4ac determines the type of zeroes of a quadratic. For p(x)=x²+2x−8, D=2²−4(1)(−8)=4+32=36. Since 36 is positive and a perfect square, p has two distinct rational real zeroes; indeed, they are 2 and −4. For q(x)=x²+2x−7, D=2²−4(1)(−7)=4+28=32. It is positive but not a perfect square, so q has two distinct irrational real zeroes. Therefore option A is correct. Option B incorrectly calls the second pair rational, option D reverses the classifications, and option C mistakes a positive discriminant for a non-real result.

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).

Core idea: compute \(p(\sqrt{7}+1)=(\sqrt{7}+1)^2-7\) by expanding. \((\sqrt{7}+1)^2=(\sqrt{7})^2+2\cdot\sqrt{7}\cdot1+1^2=7+2\sqrt{7}+1\). Subtracting 7 gives \(7+2\sqrt{7}+1-7=1+2\sqrt{7}\). Thus the correct value is \(1+2\sqrt{7}\). Closest distractor: option B (\(8+2\sqrt{7}\)) arises from forgetting to subtract the final \(7\). Option C (\(1\)) ignores the surd term \(2\sqrt{7}\). Option D (\(7+2\sqrt{7}\)) reflects a mis-evaluation of the constant terms. Exam tip: always expand \((a+b)^2\) carefully and simplify each term step by step to avoid dropping surd or constant terms.

Substitution gives (2+k\sqrt{2}-2=k\sqrt{2}=0), so (k=0). In exams write (p(\alpha)=0) for a zero.

(\frac{1}{1+\sqrt{2}}=\sqrt{2}-1), so the sum is (2\sqrt{2}). Rationalising the denominator is a quick exam method.

Because ((3+\sqrt{8})(3-\sqrt{8})=9-8=1), the reciprocal is (3-\sqrt{8}). If the product is (1), the reciprocal is immediate.

(\alpha\beta=(\sqrt{5}+2)(\sqrt{5}-2)=5-4=1). Conjugate multiplication often gives a rational answer.

The sum is (10) and the product is (25-6=19). In exams quickly find sum and product for conjugate zeroes.

The discriminant is (100-76=24), so the zeroes are (\frac{10\pm\sqrt{24}}{2}=5\pm\sqrt{6}). Simplify (\sqrt{24}=2\sqrt{6}) in exams.

The discriminant is (36-4=32), positive but not a perfect square. So there are two distinct irrational real roots.

The discriminant is (16-16=0), so the roots are equal and (x=2) is rational. In exams (D=0) means equal roots.

The discriminant is (4-20=-16), which is negative, so there are no real roots. In exams do not treat a negative discriminant as real zeroes.