Roots of a Quadratic Equation

Here (\alpha+\beta=3) and (\alpha\beta=1). First (\alpha^2+\beta^2=7), then (\alpha^4+\beta^4=7^2-2=47).

By Vieta’s relations, the sum of the roots is \(12\), while their product is \(m\). The only pair of prime numbers with sum \(12\) is \(5\) and \(7\), since \(5+7=12\). Therefore, \(m=5\times7=35\). In such questions, identify the root sum first and then use the product relation.

For equal roots, (p^2-144=0), so (p=\pm12). The equal root (-\frac{p}{2}) must be positive, hence (p=-12).

Use ((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta). With (\alpha+\beta=\frac{13}{3}) and (\alpha\beta=\frac{4}{3}), the positive difference is (\frac{11}{3}).

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).

Here (\alpha+\beta=6) and (\alpha\beta=5). Since (\alpha^2+\beta^2=26), the value is (26-5(\alpha+\beta)=-4).

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).