Irrational numbers and real numbers

(p(x)=(x-a)^2-7), so (x=a\pm\sqrt{7}). Recognizing a perfect-square form saves time in hard questions.

Use the discriminant: \(D=b^2-4ac=36-4k=4(9-k)\). For real roots we need \(D\ge0\) (i.e. \(k\le9\)). For the roots to be irrational we need \(D>0\) and \(D\) not a perfect square.

- For \(k=7\), \(D=36-28=8\), which is positive and not a perfect square, so the roots are real and irrational (correct).
- For \(k=5\), \(D=16\) which is a perfect square, giving rational roots.
- For \(k=9\), \(D=0\) so the roots are equal and rational.
- For \(k=10\), \(D=-4\) is negative, so the roots are complex (not real).

Exam tip: Always check that \(D>0\) and then test whether \(D\) is a perfect square to decide irrational vs rational roots.

The constant term is the product, and ((4+\sqrt{11})(4-\sqrt{11})=16-11=5). In conjugate products, the irrational middle part cancels.

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.

(\alpha+\beta=10) and (\alpha\beta=25-6=19), so (\alpha^2+\beta^2=100-38=62). Use (\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta).

The zeroes are (5\pm2\sqrt{2}), so the difference is (4\sqrt{2}). For conjugate zeroes, the difference is twice the radical part.

After removing the common factor, we get (x^2-6x+7), and (D=36-28=8). Since (D) is positive and not a perfect square, the zeroes are real irrational.

(\sqrt{12}=2\sqrt{3}), so the sum is (3\sqrt{3}). In a monic polynomial, the coefficient of (x) is the negative of the sum of zeroes.

For zeros \(3+\sqrt{2}\) and \(3-\sqrt{2}\) the sum is \((3+\sqrt{2})+(3-\sqrt{2})=6\) and the product is \((3+\sqrt{2})(3-\sqrt{2})=9-2=7\). A monic quadratic with these zeros is \(x^2-(\text{sum})x+(\text{product})=x^2-6x+7\). The closest distractor \(x^2-6x+11\) (option D) has the same linear coefficient but the constant term (product) is incorrect. Exam tip: form \((x-(3+\sqrt{2}))(x-(3-\sqrt{2}))\) and expand — this avoids sign mistakes.

The sum is (\sqrt{5}+\sqrt{7}) and the product is (\sqrt{35}). Both match (\sqrt{5}) and (\sqrt{7}).

(p(x)=(x-\sqrt{10})^2), so the zero (\sqrt{10}) occurs twice. A perfect-square form quickly gives equal zeroes.

(\alpha+\beta=4) and (\alpha\beta=-1), so (\frac{1}{\alpha}+\frac{1}{\beta}=\frac{4}{-1}=-4). Find sum and product first.

For (x^2-8x+3), (D=64-12=52), positive and not a perfect square. The other options give equal rational, non-real, or rational zeroes.

(\alpha+\beta=2) and (\alpha\beta=-1), so (\alpha^3+\beta^3=2^3-3(-1)(2)=14). Use (\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)).

For conjugate irrational roots the sum and product are easy: sum = \((7+2\sqrt{3})+(7-2\sqrt{3})=14\), product = \((7+2\sqrt{3})(7-2\sqrt{3})=49-(2\sqrt{3})^2=49-12=37\). For the monic quadratic \(x^2+px+q\), sum of roots = \(-p\) and product = \(q\). Hence \(p=-14\), \(q=37\) and \(p+q=37-14=23\). Closest distractor explanation: 37 is just the product \(q\), not \(p+q\); 49 would be wrong if one forgets to subtract \((2\sqrt{3})^2\). Exam tip: For a monic quadratic use sum = \(-p\), product = \(q\); compute \(p+q\) as \(q-\text{(sum of roots)}\) for speed.

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.

Substitution gives (5+a\sqrt{5}-5=0), so (a\sqrt{5}=0) and (a=0). Simplify like terms first while substituting.

By the formula, (x=\frac{8\pm\sqrt{64-8}}{4}=2\pm\frac{\sqrt{14}}{2}). Divide the whole numerator by the denominator carefully.

(\alpha+\beta=2) and (\alpha\beta=-4), so (\alpha^2+\beta^2=2^2-2(-4)=12). Symmetric values can be found without finding the zeroes.

For a monic quadratic x²+bx+c, the product of its zeroes equals the constant term c. The given zeroes are 1+√6 and 1−√6, so use the difference-of-squares identity: (1+√6)(1−√6)=1²−(√6)²=1−6=−5. Therefore m, the constant term, is −5. The sum provides an additional check: (1+√6)+(1−√6)=2, and the coefficient of x is −2, exactly the negative of that sum. Thus option A is correct. Option B loses the negative sign. The values 7 and −7 may result from incorrectly adding the radical terms or mishandling the product; neither agrees with the coefficient relations. Both Vieta’s product and sum confirm the answer.