Roots of a Quadratic Equation

The governing concept is recognising a quadratic as a difference of squares. We rewrite x^2−2px+p^2−9 as (x−p)^2−3^2, which factors into (x−p−3)(x−p+3). Hence the roots are p+3 and p−3. Their difference, taking the larger root minus the smaller root, is (p+3)−(p−3)=6 for every value of p. Therefore option A is correct. Their sum is 2p, not always 6, and their product is p^2−9, not always 9. The roots are distinct because their difference is 6, so option D is also false.

The sum (-4) is already negative and the product must be positive, so (c>0). For real roots, (16-4c\ge0), hence (0<c\le4).

We use ((\alpha+3)(\beta+3)=\alpha\beta+3(\alpha+\beta)+9). Since (\alpha+\beta=2) and (\alpha\beta=-8), the value is (7).

The sum is (\frac{3t+1}{2}) and the product is (\frac{t(t+1)}{2}). These match (-\frac{b}{a}) and (\frac{c}{a}) for every (t).

Putting (x=3) gives (9-3(m-2)+m-6=0), so (m=\frac{9}{2}). The product is (-\frac{3}{2}), so the other root is (-\frac{1}{2}); hence no option is correct.

Putting (x=3) makes the equation true for every (m). The product is (3m) and one root is (3), so the other root is (m).

Use the relationship between roots and coefficients, or factor directly. The expression x^2−(2a+1)x+a(a+1) equals (x−a)(x−a−1), because the sum of a and a+1 is 2a+1 and their product is a(a+1). Therefore the equation has roots x=a and x=a+1, making option A correct. Option B incorrectly treats the coefficient of x as a root. The roots are not always equal; they differ by 1 for every a. Since both roots are real whenever a is real, option D is also incorrect. The discriminant is 1, confirming two distinct real roots.

We use (\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}). Here (\alpha^2+\beta^2=45) and (\alpha\beta=18), so the value is (\frac{5}{2}).

We would need (\sin \theta+\cos \theta=4), but its maximum is (\sqrt{2}). Therefore no real (\theta) is possible.

By Vieta’s formula, the product of the roots of \(x^2-2x+k=0\) is \(k\). Here, the product is \(\tan\theta\cdot\cot\theta=1\), so \(k=1\). The value 2 is related to the sum of the roots, not their product. Exam tip: for a monic quadratic \(x^2+bx+c=0\), the product of the roots is \(c\).

Since \((2+\sqrt{3})(2-\sqrt{3})=1\), the given reciprocals are \(2-\sqrt{3}\) and \(2+\sqrt{3}\). Their sum is \(4\). For \(x^2+ax+b=0\), the sum of the roots is \(-a\). Thus, \(-a=4\), giving \(a=-4\). Exam tip: For a quadratic equation, use the sum-of-roots relation \(-\frac{\text{coefficient of }x}{\text{coefficient of }x^2}\).

By Vieta’s formula, the product of the roots of \(Ax^2+Bx+C=0\) is \(C/A\). Here, \(A=1\) and \(C=a^2-1\), so the product of the roots is \(a^2-1\). Setting it equal to zero gives \(a^2-1=0\), or \(a^2=1\), hence \(a=1\) or \(a=-1\). Therefore, option A is correct. Exam tip: Use Vieta’s formulas directly when a question asks for the sum or product of roots, and remember to consider both values obtained after solving the parameter equation.

By Vieta’s formulas, (\alpha+\beta=6) and (\alpha\beta=n). Using (\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta), we get (20=6^2-2n=36-2n). Hence, (2n=16) and (n=8). Exam tip: For the sum of squares of roots, use ((\alpha+\beta)^2-2\alpha\beta) rather than finding the roots individually.

If the roots are \(\alpha\) and \(\beta\) and are reciprocals, then \(\alpha\beta=1\). By Vieta’s relation, \(\alpha\beta=c/a\), so \(c/a=1\Rightarrow c=a\). Coefficient \(b\) relates to the sum of roots. Exam tip: for reciprocal roots, check the product first.

Let the roots be (2r) and (3r). From (5r=12), (r=\frac{12}{5}), so the product is (6r^2=\frac{864}{25}).

Factoring the polynomial gives \(x^2-(a+b)x+ab=(x-a)(x-b)\). Hence, by the zero-product property, \((x-a)(x-b)=0\) implies \(x=a\) or \(x=b\). Therefore, the roots are \(a\) and \(b\). Option B lists the sum and product of the roots, not the roots themselves. Exam tip: In \(x^2-Sx+P=0\), look for two numbers whose sum is \(S\) and product is \(P\).

We use (\alpha^2\beta+\alpha\beta^2=\alpha\beta(\alpha+\beta)). Here (\alpha+\beta=-5), so (r(-5)=-20) and (r=4).

Here (\alpha+\beta=4) and (\alpha\beta=1). First (\alpha^2+\beta^2=14), then (\alpha^4+\beta^4=14^2-2=194).

The prime pairs with sum (10) are ((3,7)) and ((5,5)). Hence (m=21) or (m=25), and their sum is (46), so option (B) should be correct.

For equal roots, (p^2-64=0), so (p=\pm8). The root (-\frac{p}{2}) must be positive, hence (p=-8).