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.

(\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.

Direct answer: option A, \(a=0\). Since \(p(\sqrt5)=0\) and \(p(x)=x^2+ax-5\), substitute \(x=\sqrt5\): \((\sqrt5)^2+a\sqrt5-5=0\). This becomes \(5+a\sqrt5-5=0\), so \(a\sqrt5=0\). Because \(\sqrt5\neq0\), division by \(\sqrt5\) gives \(a=0\). Option A is correct. Option B, \(\sqrt5\), would make the middle term 5 and not zero. Option C, \(-\sqrt5\), would make the middle term -5. Option D, 5, would make the middle term \(5\sqrt5\). The important step is to square \(\sqrt5\) correctly as 5 before comparing terms. Memory cue: substitute first, simplify the square, then solve for the coefficient.

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.

(\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.

Direct answer: A, \(-2+\sqrt2\) and \(-2-√2\). Use the quadratic formula for \(x^2+4x+2=0\): \(x=\frac{-4\pm\sqrt{4^2-4(1)(2)}}{2(1)}=\frac{-4\pm
u000b√8}{2}=\frac{-4\pm2\sqrt2}{2}=-2\pm\sqrt2\). Therefore the two zeroes are exactly those in A. Option B has the wrong sign before 2 and would have positive sum 4. Option C uses \(-4\) instead of the required halved value \(-2\). Option D uses \(\sqrt6\), but the discriminant gives \(\sqrt8=2\sqrt2\), not \(\sqrt6\). A quick check is useful: the sum of A is \(-4\), and the product is \((-2)^2-(\sqrt2)^2=2\), matching the polynomial. Common mistake: forgetting to divide every term in the numerator by 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.