Roots of a Quadratic Equation

The product of roots is (\frac{k}{k}=1), so (k\neq0) is needed. For real roots, (100-4k^2\ge0), hence (k^2\le25).

The direct answer is option B: 8 and −8. For x²+px+15=0, the product of the roots is the constant term 15. The roots are r and r+2, so r(r+2)=15. Expanding gives r²+2r−15=0, which factorises as (r+5)(r−3)=0. Therefore r=3 or r=−5. If r=3, the roots are 3 and 5, whose sum is 8. If r=−5, the roots are −5 and −3, whose sum is −8. For x²+px+15, the sum of roots is −p. Thus p=−8 in the first case and p=8 in the second case. The possible values are 8 and −8, so option B is correct. Option A, 6 and −6, does not come from the required root pairs. Option C, 10 and −10, confuses the possible root sums with another value and is not obtained here. Option D, 15 and −15, is related to the constant/product, not the coefficient p. Remember: product of roots is c/a, while sum is −b/a.

A quadratic equation has real roots when its discriminant satisfies \(D\ge0\). Here, \(A=9\), \(B=-6(a+1)\), and \(C=a^2-3a\). Thus, \(D=B^2-4AC=36(a+1)^2-36(a^2-3a)=36(5a+1)\). Therefore, \(36(5a+1)\ge0\), giving \(a\ge-\frac{1}{5}\). Option B reverses the inequality and incorrectly excludes the boundary value \(a=-\frac{1}{5}\). Exam tip: for real roots, use \(D\ge0\); the case \(D=0\) represents equal real roots and is included.

By Vieta’s formulas, \(\alpha+\beta=7\) and \(\alpha\beta=12\). Combining the fractions gives \(\frac{(\alpha+1)(\beta-1)+(\beta+1)(\alpha-1)}{(\alpha-1)(\beta-1)}=\frac{2\alpha\beta-2}{\alpha\beta-(\alpha+\beta)+1}\). Therefore, the value is \(\frac{2(12)-2}{12-7+1}=\frac{22}{6}=\frac{11}{3}\). Hence, option C is correct; an answer such as \(\frac{10}{3}\) results from an error while simplifying the numerator or denominator. Exam tip: In such problems, use the sum and product of the roots instead of finding the roots individually.

Here (\alpha+\beta=\frac{8}{3}) and (\alpha\beta=\frac{4}{3}). Using (\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)), we get (\frac{224}{27}).

For equal roots, the discriminant \(D=b^2-4ac\) must be zero. Here, \(a=1\), \(b=m+1\), and \(c=16\). Thus, \((m+1)^2-4(1)(16)=0\), giving \((m+1)^2=64\). Therefore, \(m+1=\pm 8\), so \(m=7\) or \(m=-9\). Hence, option A is correct. Exam tip: For equal roots of a quadratic equation, set \(b^2-4ac=0\).

By Vieta’s formulas, the sum of the roots is \(-a\) and their product is \(b\). Here, \((3+\sqrt{2})+(3-\sqrt{2})=6\), so \(a=-6\). Also, \((3+\sqrt{2})(3-\sqrt{2})=9-2=7\), so \(b=7\). Therefore, \(a+b=-6+7=1\). Exam tip: In \(x^2+ax+b=0\), the sum of the roots is \(-a\), not \(a\); check this sign carefully.

Since (\alpha) is a root, (\alpha^2=4\alpha+2), and (\alpha\beta=-2). The expression becomes (4\alpha+2+4\beta-2=4(\alpha+\beta)=16).

The equation can be rewritten as \((x-q)^2-16=0\), so \((x-q)^2=16\). Hence, the roots are \(x=q+4\) and \(x=q-4\). Their positive difference is \((q+4)-(q-4)=8\). Exam tip: when a quadratic is in the form \((x-a)^2=b^2\), its roots are \(a+b\) and \(a-b\), making their difference \(2b\).

The sum (5) is positive and product (c>0) is needed for both roots. For real roots, (25-4c\ge0), so (0<c\le\frac{25}{4}).

We use ((\alpha-4)(\beta-4)=\alpha\beta-4(\alpha+\beta)+16). Since (\alpha+\beta=3) and (\alpha\beta=-10), the value is (-6).

The sum of these two roots is (\frac{4t+2}{3}), and the product is (\frac{t(t+2)}{3}). These match (-\frac{b}{a}) and (\frac{c}{a}) of the given equation.

Factoring the equation gives x² − 2(a + 1)x + a² + 2a = (x − a)(x − a − 2). Therefore, the roots are x = a and x = a + 2. In option B, both roots would be a + 1; although their sum is correct, their product would be (a + 1)², not a² + 2a. Exam tip: Verify the roots using the sum 2(a + 1) and the product a² + 2a.

We use (\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}). Here (\alpha^2+\beta^2=73) and (\alpha\beta=24), so the value is (\frac{73}{24}).

We have (\tan\theta\cdot\cot\theta=1) and (\tan\theta+\cot\theta=s). For real values, (s^2-4\ge0), so (s^2\ge4).

After rationalising, the sum of roots is (\frac{3}{2}). In (x^2+ax+b=0), the sum is (-a), so (a=-\frac{3}{2}).

By Vieta’s formula, the product of the roots of \(Ax^2+Bx+C=0\) is \(C/A\). Here, \(A=1\) and \(C=a^2-9\), so the product of the roots is \(a^2-9\). Setting it equal to zero gives \(a^2-9=0\), or \(a^2=9\), hence \(a=3\) or \(a=-3\). Option A is incorrect because \(a^2=9\) does not give 0 and 9 as the values of \(a\). Exam tip: Use Vieta’s formulas directly when only the sum or product of roots is given.

By Vieta’s formulas, \(\alpha+\beta=8\) and \(\alpha\beta=n\). Using \(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta\), we get \(40=8^2-2n=64-2n\). Thus, \(2n=24\) and \(n=12\). Exam tip: when the sum of the squares of the roots is given, use \((\alpha+\beta)^2-2\alpha\beta\) directly.

Let the roots be 3r and 4r. Their sum is 14, so 3r + 4r = 14, giving r = 2. Thus, the roots are 6 and 8. In the quadratic equation x² − 14x + q, the product of the roots equals q; hence q = 6 × 8 = 48. Exam tip: For x² + bx + c, the sum of the roots is −b and their product is c.

The polynomial \(x^2-(u-v)x-uv\) can be factorised as \((x-u)(x+v)\). Therefore, \(x-u=0\) or \(x+v=0\), giving the roots \(u\) and \(-v\). Their sum is \(u-v\) and their product is \(-uv\), confirming the result. In the exam, factorisation is the quickest method when the expression matches this pattern.

We use (\alpha^2\beta+\alpha\beta^2=\alpha\beta(\alpha+\beta)). Here (\alpha+\beta=6), so (6r=54) and (r=9).

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}).