Irrational numbers and real numbers

For a quadratic \(x^2+ax+c\), the sum of zeros \(\alpha+\beta=-a\) and the product \(\alpha\beta=c\). Here \(\alpha=-\sqrt{6}\) and \(\beta=\sqrt{6}\), so the sum is \(0\). Thus \(-a=0\) gives \(a=0\). Also the product \((-\sqrt{6})(\sqrt{6})=-6\) matches the constant term \(-6\), confirming consistency. The closest distractor 6 is incorrect because it would contradict the product sign; option \(-6\) is the constant term, not the coefficient \(a\). Exam tip: always use both sum and product relations to check consistency when specific roots are given — sum gives the coefficient, product checks the constant term.

The sum of the roots is \((1+\sqrt{2})+(1-\sqrt{2})=2\) and the product is \((1+\sqrt{2})(1-\sqrt{2})=1-2=-1\). For a monic quadratic with roots \(\alpha\) and \(\beta\) the polynomial is \(x^2-(\alpha+\beta)x+\alpha\beta\). Thus the required polynomial is \(x^2-2x-1\). Closest distractor \(x^2+2x-1\) has sum \(-2\) (roots \(-1\pm\sqrt{2}\)), so it does not match; \(x^2-2x+1=(x-1)^2\) has a double root 1, and \(x^2+x-2\) has different sum/product. Exam tip: compute sum and product of given roots first, then plug into \(x^2-(\text{sum})x+(\text{product})\).

Let the roots be \(\alpha=\sqrt{2}+\sqrt{3}\) and \(\beta=\sqrt{2}-\sqrt{3}\). Then \(\alpha+\beta=2\sqrt{2}\) and \(\alpha\beta=(\sqrt{2})^2-(\sqrt{3})^2=2-3=-1\). For a quadratic with these roots use \(x^2-(\text{sum})x+\text{product}\), giving \(x^2-2\sqrt{2}\,x-1\) (option A). Option B has the wrong sign on the linear term, so its sum of roots would be \(-2\sqrt{2}\); option C uses \(2\sqrt{3}\) which does not match the actual sum; option D equals \(0\) at ±\(\sqrt{5}\), not the given irrational roots. Exam tip: compute sum and product of the given roots first — this directly yields the required quadratic polynomial.

With rational coefficients, an irrational root containing a square root has its conjugate as the other root. If one zero is \\(4+\sqrt{11}\\), the other is \\(4-\sqrt{11}\\). Their sum is \\((4+\sqrt{11})+(4-\sqrt{11})=8\\). Option C (\\(2\sqrt{11}\\)) mistakes combining the radicals instead of canceling them; options B and D come from incorrect arithmetic. Exam tip: use the conjugate-root property or Vieta’s formula (sum = \(-b/a\)) to find the sum quickly.

For the quadratic, \(a=1,\ b=2,\ c=-1\). Using the quadratic formula \(x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\). Here \(b^2-4ac=4-4(1)(-1)=8\) and \(\sqrt{8}=2\sqrt{2}\). Thus \(x=\dfrac{-2\pm2\sqrt{2}}{2}=-1\pm\sqrt{2}\). Therefore the zeroes are \(-1+\sqrt{2}\) and \(-1-\sqrt{2}\). The closest distractor \(1\pm\sqrt{2}\) is wrong due to an incorrect sign (the numerator should be \(-2\pm2\sqrt{2}\), not \(+2\pm2\sqrt{2}\)); the option with \(\sqrt{3}\) is wrong because the discriminant is not 3. Exam tip: compute and simplify the discriminant first (e.g. \(\sqrt{8}=2\sqrt{2}\)) to avoid sign and simplification mistakes.

Here \(a=1,\; b=-2,\; c=-3\). The discriminant is \(D=b^2-4ac =(-2)^2-4(1)(-3)=4+12=16>0\). Hence the quadratic has two distinct real zeroes; solving gives \(x=\dfrac{2\pm\sqrt{16}}{2}\), i.e. \(x=3\) and \(x=-1\). Option B is the closest distractor — it states \(D=8\), which is an arithmetic mistake. Options C and D are also incorrect: C claims \(D<0\) (complex roots) which is false here, and D's reasoning is wrong because irrational roots are still real. Exam tip: correctly identify \(a,b,c\) and compute \(D=b^2-4ac\) to determine root nature quickly.

For real distinct zeroes, (D>0) is required. For irrational zeroes, (D) must not be a perfect square.

Remove radicals by squaring. Compute \(x^2=(\sqrt{2}+\sqrt{5})^2=2+5+2\sqrt{10}=7+2\sqrt{10}\), so \(x^2-7=2\sqrt{10}\). Squaring again gives \((x^2-7)^2=40\), hence \(x^4-14x^2+49=40\) and therefore \(x^4-14x^2+9=0\). Thus option B is correct. Option C is wrong because it would force \(2\sqrt{10}=0\). Options A and D have incorrect signs/coefficients. Exam tip: for nested radicals square to eliminate them stepwise; using conjugates also helps find the minimal polynomial over Q.

For a quadratic \(ax^2+bx+c\), the product of the zeros equals \(\frac{c}{a}\). Here \(a=1\) and \(c=2\), so the product is \(\frac{2}{1}=2\). Option B (\(2\sqrt{3}\)) is incorrect — it is the coefficient \(b\), not the product. Exam tip: remember sum of roots = \(-b/a\) and product = \(c/a\) to answer such questions quickly.

The sum is (3\sqrt{2}+\frac{3}{\sqrt{2}}=\frac{9\sqrt{2}}{2}). Hence (m=\frac{9\sqrt{2}}{2}).

(\sqrt{12}=2\sqrt{3}), so (\alpha+\beta=2\sqrt{3}-\sqrt{3}=\sqrt{3}). Simplifying radicals is important.

A polynomial P(x) has a root a if P(a)=0. Substituting \(x=\sqrt{3}\) into \(x^2-3\) gives \((\sqrt{3})^2-3=3-3=0\), so \(x^2-3\) has \(\sqrt{3}\) as a root. A tempting distractor \(x^2-6x+9\) equals \((x-3)^2\) and has root 3, not \(\sqrt{3}\). Quick exam tip: either substitute the value directly or factor the polynomial to spot the roots.

For a quadratic \(ax^2+bx+c\), the sum of zeros is \(-b/a\) and the product is \(c/a\). Here \(a=1, b=p, c=q\). The sum of the given zeros is \((7+\sqrt{13})+(7-\sqrt{13})=14\), so \(-p=14\) giving \(p=-14\). The product is \((7+\sqrt{13})(7-\sqrt{13})=49-13=36\), so \(q=36\). Therefore \(p+q=-14+36=22\). Note: \(49\) would be a mistake from taking only \(7^2\); \(35\) is a common arithmetic slip. Exam tip: identify \(a,b,c\) first and then use sum and product of roots formulas to avoid sign errors.

Use the relations between roots and coefficients for a quadratic. For \(x^2-6x+n\), sum of roots is \(-(-6)/1=6\) and product of roots is \(n\). For roots \(3+\sqrt{m}\) and \(3-\sqrt{m}\), the product is \((3+\sqrt{m})(3-\sqrt{m})=9-m\). Setting this equal to \(n=5\) gives \(9-m=5\), so \(m=4\). Option B (5) is a common distractor from confusing \(m\) with \(n\); option C (9) would give product zero, not 5. Exam tip: immediately write sum and product relations for quadratics to avoid unnecessary algebra.

This is a perfect-square trinomial because \((x-2\sqrt{2})^2 = x^2 -2\cdot(2\sqrt{2})x + (2\sqrt{2})^2 = x^2 -4\sqrt{2}x +8\). Option (x+2\sqrt{2})^2 would give +4\sqrt{2}x (wrong sign). Option (x-\sqrt{2})^2 has constant term 2, and (x-4\sqrt{2})^2 has constant term 32, so they do not match the given polynomial. Exam tip: use the identity \((x-a)^2 = x^2 -2ax + a^2\); find a from the middle coefficient and verify a^2 equals the constant term.

((a+\sqrt{5})+(a-\sqrt{5})=2a) and ((a+\sqrt{5})(a-\sqrt{5})=a^2-5). These match the polynomial.

For a quadratic \(ax^2+bx+c\) the product of the zeroes equals \(\dfrac{c}{a}\). Here \(a=1\) and \(c=-2\), so the product is \(\dfrac{-2}{1}=-2\). Option B (2) is incorrect due to wrong sign; options C (-1) and D (4) are numerically incorrect. Exam tip: remember product = \(c/a\) and always check the sign of \(c\).

Complete the square: \(p(x)=x^2-2\sqrt{7}x+7=(x-\sqrt{7})^2\). Hence the root is \(x=\sqrt{7}\), occurring twice (a double root). Using the discriminant: \(\Delta=b^2-4ac =(-2\sqrt{7})^2-4\cdot1\cdot7=28-28=0\). \(\Delta=0\) implies one real repeated root. Thus B is wrong because \(\Delta\neq>0\) so roots are not distinct; C is wrong because the root is irrational (\(\sqrt{7}\)), and D is wrong because a real root does exist. Exam tip: compute \(\Delta\) first—\(\Delta=0\) immediately indicates a repeated real root; factorization gives the exact value.

The quadratic is a perfect square: \(p(x)=(x+\sqrt{3})^2\). Expanding gives \((x+\sqrt{3})^2=x^2+2\sqrt{3}x+3\), so the only root is \(x=-\sqrt{3}\) with multiplicity two. Using the discriminant: \(\Delta=(2\sqrt{3})^2-4\cdot1\cdot3=12-12=0\), confirming a repeated real root. The closest distractor \(\sqrt{3}\) is wrong due to the sign. Exam tip: if \(\Delta=0\) or you can complete the square, expect a double root and compute \(-b/2a\).

With two zeros we treat the polynomial as a quadratic. For a monic polynomial (leading coefficient 1) the constant term equals the product of the zeros. Compute: \((2\sqrt{2})(3\sqrt{2})=2\cdot3\cdot(\sqrt{2}\cdot\sqrt{2})=6\cdot2=12\). Option C (\(6\sqrt{2}\)) is a common mistake where the factor \(\sqrt{2}\cdot\sqrt{2}=2\) was not applied. Exam tip: For a monic polynomial of degree \(n\), the constant term = \((-1)^n\) times the product of the roots; for quadratics this equals the product directly.