Roots of a Quadratic Equation

The polynomial factors as \((x-(a+3))(x-(a+4))=0\), since its expansion is \(x^2-(2a+7)x+(a+3)(a+4)\). Therefore, the roots are \(a+3\) and \(a+4\), whose difference is 1, so they are consecutive integers. Option A has the correct sum of roots but not the correct product. Exam tip: Factor the quadratic first, then verify the roots using their sum and product.

For both roots to be negative, the sum (-10) and product (\lambda>0) are needed. For real distinct roots, (100-4\lambda>0), hence (0<\lambda<25).

The direct answer is B: \(-4\). By Vieta’s formulas for \(x^2-6x+5=0\), \(\alpha+\beta=6\) and \(\alpha\beta=5\). We need \(\alpha^2-5\alpha+\beta^2-5\beta\). Grouping terms gives \(\alpha^2+\beta^2-5(\alpha+eta)\). Now \(\alpha^2+\beta^2=(\alpha+eta)^2-2\alpha\beta=6^2-2(5)=26\). Therefore the expression is \(26-5(6)=26-30=-4\). Option A, -6, results from an arithmetic or formula error. Option B is correct. Option C, 4, has the wrong sign. Option D, 6, ignores the subtraction term. Memory cue: convert separate root expressions into sums and products before calculating.

By Vieta’s formulas, \(\alpha+\beta=-8\) and \(\alpha\beta=12\). Therefore, \((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta=(-8)^2-4(12)=64-48=16\). Hence, \((\alpha-\beta)^2+3\alpha\beta=16+3(12)=52\). Exam tip: For expressions involving the sum and product of roots, apply Vieta’s formulas directly; using only \((\alpha+\beta)^2\) leads to an incorrect result.

By Vieta’s formulas, for \(x^2+px+q=0\), the sum of the roots is \(-p\) and their product is \(q\). Here, \((-3)+7=4\), so \(p=-4\), while \((-3)(7)=-21\), so \(q=-21\). Therefore, \(p-q=-4-(-21)=17\). Exam tip: in a monic quadratic \(x^2+px+q\), the sum of the roots is \(-p\), not \(p\).

For a quadratic equation \(ax^2+bx+c=0\), the product of the roots is \(\frac{c}{a}\). Here, \(a=3\) and \(c=h\), so the product of the roots is \(\frac{h}{3}\). Since one root is \(\frac{1}{3}\), the other root is \(\frac{h/3}{1/3}=h\). Option B is the product of the roots, not the second root. Exam tip: To find an unknown root, divide the product of the roots by the known root.

The discriminant is (D=4-4(a^2+2)=-4(a^2+1)). It is negative for every real (a), so the roots are not real.

Here (\alpha+\beta=5) and (\alpha^2+\beta^2=17). From (25-2\alpha\beta=17), (\alpha\beta=4), so the roots are (1) and (4).

For equal roots, (b^2-196=0), so (b=\pm14). The equal root (-\frac{b}{2}) must be negative, hence (b=14).

Factorising the equation gives \(x^2+x-2=(x-1)(x+2)\), so its roots are \(1\) and \(-2\). Therefore, \(\alpha^5+\beta^5=1^5+(-2)^5=1-32=-31\). The distractor \(31\) results from missing the negative sign of the odd power of \(-2\). Exam tip: an odd power of a negative number remains negative.

The polynomial can be factorised as \((x-(u+1))(x-(v+1))=0\). Therefore, its roots are \(u+1\) and \(v+1\). If the roots are equal, then \(u+1=v+1\), which gives \(u=v\). Options A, B and D are not necessary conditions for equal roots. Exam tip: In such questions, comparing the roots after factorisation is usually quicker than calculating the discriminant.

Here (\alpha+\beta=5) and (\alpha\beta=6). The new roots are (5) and (6), so the equation is (x^2-11x+30=0).

The equation can be rewritten as \\((x-k)^2-25=0\\). Thus, \\((x-k)^2=25\\), giving the roots \\(k+5\\) and \\(k-5\\). Their positive difference is \\((k+5)-(k-5)=10\\), so option B is correct. Exam tip: For \\(ax^2+bx+c=0\\), the difference between the roots can also be found using the discriminant.

For real roots, the discriminant must satisfy \(D\ge 0\). Here \(a=1\), \(b=-2m\), and \(c=m^2-m\), so \(D=b^2-4ac=(-2m)^2-4(m^2-m)=4m\). Therefore, \(4m\ge 0\), giving \(m\ge 0\). Option A is incorrect because \(m=0\) also gives two equal real roots. Exam tip: For questions about real roots, first apply the discriminant condition.

If the roots are \(\alpha\) and \(\beta\), the monic quadratic equation is \(x^2-(\alpha+\beta)x+\alpha\beta=0\). Substituting \(\alpha+\beta=7\) and \(\alpha\beta=10\) gives \(x^2-7x+10=0\). Option A has the wrong sign for the sum, while option B interchanges the sum and product. Exam tip: in the monic form, the coefficient of \(x\) is the negative of the sum of roots, and the constant term is their product.

Reciprocal roots have product (1). Multiplying (x^2-\frac{5}{2}x+1=0) by (2) gives (2x^2-5x+2=0).

The product of roots is (\frac{m-1}{m-1}=1), so (m\neq1) is needed. For real roots, (D=16m\ge0), hence (m\ge0) and (m\neq1).

The equation becomes ((x-(a+1))(x-(a+5))=0). So the roots are (a+1) and (a+5), hence the positive difference is (4).

Here (\alpha+\beta=\frac{16}{5}) and (\alpha\beta=\frac{p}{5}). Using ((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta), we get (p=11).

For equal roots, put (D=0). From (4(a+5)^2-4(a^2+18)=0), we get (10a+7=0), so (a=-\frac{7}{10}).

Factor the original quadratic: x^2 - 7x + 12 = (x - 3)(x - 4), so alpha and beta are 3 and 4. Applying the stated transformation to each root gives 2alpha + 3 = 2(3) + 3 = 9 and 2beta + 3 = 2(4) + 3 = 11. We now need the quadratic whose roots are 9 and 11.

For roots r and s, the equation is x^2 - (r+s)x + rs = 0. Their sum is 9 + 11 = 20 and their product is 9 × 11 = 99. Therefore the required equation is x^2 - 20x + 99 = 0, so option A is correct. The coefficient -14 in option B does not use the transformed-root sum, and the other choices have an incorrect sign or constant term.

Let the roots be (2r) and (5r). From (10r^2=\frac{10}{3}), (r=\pm\frac{1}{\sqrt{3}}), so (7r=-\frac{m}{3}) gives (m=\pm7\sqrt{3}).

Here (\alpha^2+\beta^2=64-4=60) and ((\alpha\beta)^2=4). Thus (\frac{1}{\alpha^2}+\frac{1}{\beta^2}=\frac{60}{4}=15).

The product of roots is (\frac{k}{k}=1), so (k\neq0) is needed. For real roots, (144-4k^2\ge0), hence (k^2\le36).

From (r(r+2)=24), we get (r=4) or (r=-6). The sum of roots is (10) or (-10), so (p=-10) or (p=10).