Irrational numbers and real numbers

For the monic quadratic \(x^2-10x+k\), the product of the zeros equals the constant term \(k\). Compute the product: \((5+\sqrt{2})(5-\sqrt{2})=5^2-(\sqrt{2})^2=25-2=23\). Hence \(k=23\). The nearby option 25 is incorrect because it is simply \(5^2\), not the actual product of the conjugate roots. Exam tip: remember for \(ax^2+bx+c\), sum of roots = \(-b/a\) and product = \(c/a\); for a monic quadratic product = constant term.

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.

Direct answer: option A, so \(m=\frac{9\sqrt{2}}{2}\). For a quadratic \(x^2-mx+9\), the sum of zeroes is \(m\), because the coefficient rule gives \(\alpha+\beta=-\frac{-m}{1}=m\). Therefore add the given zeroes: \(3\sqrt{2}+\frac{3}{\sqrt{2}}\). Rationalising the second term, \(\frac{3}{\sqrt{2}}=\frac{3\sqrt{2}}{2}\), so the sum is \(\frac{6\sqrt{2}}{2}+\frac{3\sqrt{2}}{2}=\frac{9\sqrt{2}}{2}\). Option A matches. B, 9, is not this sum. C is only one zero, not both added. D has an incorrect factor and is not the sum. Memory cue: in a monic quadratic \(x^2+bx+c\), sum of zeroes is \(-b\).

(\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.

Direct answer: A. For a quadratic \(x^2+bx+c=0\), the sum of zeroes is \(-b\) and their product is \(c\). Here the polynomial is \(x^2-2ax+(a^2-5) imes1=0\), so the sum should be \(2a\), and the product should be \(a^2-5\). Directly, the proposed zeroes give \((a+\sqrt5)+(a-√5)=2a\). Their product is \((a+\sqrt5)(a-√5)=a^2-5\), using difference of squares. Thus A matches both required relationships. B gives the wrong sum and product. C also gives incorrect relationships. D would describe different roots, not these two; in particular their sum is not generally zero and their product is not generally \(-5\). The safest check is to add the roots and multiply them, then compare with the coefficients. Memory cue: for a monic quadratic, roots \(r,s\) produce \(x^2-(r+s)x+rs\).

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.

For a monic quadratic the relation is \(x^2-Sx+P\), where \(S\) is the sum of the roots and \(P\) their product. With \(S=5\sqrt{2}\) and \(P=12\) the polynomial is \(x^2-5\sqrt{2}x+12\). Option B has the wrong sign on the linear term; for \(x^2+5\sqrt{2}x+12\) the sum of roots would be \(-5\sqrt{2}\), not \(+5\sqrt{2}\). Options C and D do not match the given sum and product either. Exam tip: Use \(x^2-(\text{sum})x+(\text{product})\) for monic quadratics and double-check signs of coefficients.

For a quadratic \(ax^2+bx+c\), the product of the zeroes equals \(\frac{c}{a}\). Here \(a=1\) and \(c=\sqrt{2}\), so the product is \(\frac{\sqrt{2}}{1}=\sqrt{2}\). Option B (\(-\sqrt{2}\)) is wrong because the signs of \(a\) and \(c\) are both positive, giving a positive product. Exam tip: always identify \(a,b,c\) first and use \(\frac{c}{a}\) for the product and \(-\frac{b}{a}\) for the sum of roots.

The direct answer is option A: the sum of the zeroes is \(2\sqrt{2}\). For a quadratic polynomial \(ax^2+bx+c\), the sum of its zeroes is \(-b/a\). Here, \(a=\sqrt{2}\) and \(b=-4\). Therefore, sum \(=-(-4)/\sqrt{2}=4/\sqrt{2}\). Rationalising or simplifying gives \(4/\sqrt{2}=4\sqrt{2}/2=2\sqrt{2}\). Option A is exactly this value. Option B gives \(\sqrt{2}/\sqrt{2}=1\), so it is too small. Option C gives \(4\sqrt{2}/4\sqrt{2}=1\), also incorrect. Option D gives \(2/2=1\), incorrect as well. Remember: for \(ax^2+bx+c\), add the roots using \(-b/a\), not \(-c/a\).

For a quadratic ax^2+bx+c the product of zeroes is \(\alpha\beta=\frac{c}{a}\). Here a=\(\sqrt{3}\) and c=\(2\sqrt{3}\), so \(\alpha\beta=\frac{2\sqrt{3}}{\sqrt{3}}=2\). Option B (\(\sqrt{3}\)) is incorrect — it results from mishandling the radical cancellation or using wrong coefficients. Exam tip: Use \(\frac{c}{a}\) directly and cancel common radical factors carefully to avoid mistakes.