Roots of a Quadratic Equation

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.

The direct answer is A: \(q=864/25\). Let the roots be \(2r\) and \(3r\), because their ratio is 2:3. For \(x^2-12x+q=0\), the sum of roots is 12. Therefore \(2r+3r=5r=12\), so \(r=12/5\). Their product is \(q\), since the coefficient of \(x^2\) is 1. Thus \(q=(2r)(3r)=6r^2=6(144/25)=864/25\). Option A is correct. Option B is only 72/5 and does not equal the product. Option C is half the required product. Option D, 36, would correspond to roots whose product is 36, not roots in the stated ratio with sum 12. Use the sum first, then calculate the product.

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).

Use ((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta). With (\alpha+\beta=\frac{7}{2}) and (\alpha\beta=\frac{3}{2}), the positive difference is (\frac{5}{2}).

Factoring the constant term gives \(a^2+3a+2=(a+1)(a+2)\), and the sum of these factors is \((a+1)+(a+2)=2a+3\). Thus, the equation can be written as \([x-(a+1)][x-(a+2)]=0\), so its roots are \(a+1\) and \(a+2\). For them to be consecutive integers, \(a\) must be an integer. Option B has the correct sum, but its product is \(a^2+3a\), which lacks the required \(+2\). Exam tip: Always verify both the sum and product of the proposed roots.

For both roots to be negative, the sum (-2) and product (\lambda>0) are needed. For real distinct roots, (4-4\lambda>0), hence (0<\lambda<1).

(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=13). Thus the value is (13-4(5)=-7), so none of the options is correct.

Here (\alpha+\beta=5) and (\alpha\beta=6). Since (\alpha^2+\beta^2=13), the value is (13-4(\alpha+\beta)=13-20=-7).

By Vieta’s formulas, \(\alpha+\beta=-6\) and \(\alpha\beta=8\). Therefore, \((\alpha-eta)^2=(\alpha+eta)^2-4\alpha\beta=36-32=4\). Hence, \((\alpha-eta)^2+2\alpha\beta=4+2(8)=20\), so option A is correct. Exam tip: In such questions, use the sum and product of the roots directly instead of finding the roots individually.

By Vieta’s formulas, the sum of the roots is \(-p\) and their product is \(q\). Here, \((-2)+5=3\), so \(-p=3\), giving \(p=-3\). Also, \((-2)(5)=-10\), so \(q=-10\). Therefore, \(p-q=-3-(-10)=7\). Exam tip: for an equation of the form \(x^2+px+q=0\), remember that the sum of roots is \(-p\) and the product is \(q\).

The product is (\frac{h}{2}). Since one root is (\frac{1}{2}), the other root is (\frac{h}{2}\div\frac{1}{2}=h).

The discriminant is \(D=b^2-4ac=(-2)^2-4(a^2+1)=-4a^2\). When \(a=0\), \(D=0\), and the equation becomes \((x-1)^2=0\), giving two equal real roots. When \(a\ne0\), \(D<0\), so the equation has no real roots. Therefore, option A is correct; option C is incorrect because equal roots occur only when \(a=0\). Exam tip: To determine the nature of quadratic roots, first examine the sign of the discriminant.

Here (\alpha+\beta=4) and (\alpha^2+\beta^2=10). From (16-2\alpha\beta=10), (\alpha\beta=3), so the roots are (1) and (3).

For equal roots, (b^2-100=0), so (b=\pm10). The equal root (-\frac{b}{2}) must be negative, hence (b=10).

Here (\alpha+\beta=-1) and (\alpha\beta=-1). The power sums give (S_1=-1), (S_2=3), (S_3=-4), (S_4=7), (S_5=-11).

The equation \(x^2-(u+v)x+uv=0\) factors as \((x-u)(x-v)=0\), so its roots are \(u\) and \(v\). For the roots to be equal, it is necessary that \(u=v\). Equivalently, the discriminant is \(D=(u+v)^2-4uv=(u-v)^2\); equal roots require \(D=0\), giving \(u=v\). Exam tip: Look for the factorised form first, or use the discriminant condition directly.

In the given equation, the sum of roots is (2r+5) and the product is (r^2+5r+6=(r+2)(r+3)). Hence the roots are (r+2) and (r+3), so the positive difference is (1).

Here (\alpha+\beta=3) and (\alpha\beta=\frac{m}{4}). Putting ((\alpha-\beta)^2=1) gives (9-m=1), so (m=8).

For equal roots, the discriminant must be zero: \(D=b^2-4ac=0\). Here, \(a_q=1\), \(b=-2(a-4)\), and \(c=a^2-20\). Therefore, \(4(a-4)^2-4(a^2-20)=0\), which simplifies to \(144-32a=0\), giving \(a=\frac{9}{2}\). Hence, option B is correct. Exam tip: For equal roots, immediately use \(D=0\), and distinguish the parameter \(a\) from the standard quadratic coefficient usually denoted by \(a_q\).

The original equation can first be factored to identify its roots. Since x^2 - 9x + 14 = (x - 2)(x - 7), its roots are alpha = 2 and beta = 7. The requested new roots are obtained by subtracting 3 from each original root, so they are -1 and 4. A quadratic with roots r and s is x^2 - (r+s)x + rs = 0.

For the new roots, the sum is -1 + 4 = 3 and the product is (-1)(4) = -4. Thus the required equation is x^2 - 3x - 4 = 0, which is option A. The original equation itself would still have roots 2 and 7, not the transformed roots. The signs in the sum and product confirm the stated answer.

We use (\frac{1}{\alpha^2}+\frac{1}{\beta^2}=\frac{\alpha^2+\beta^2}{(\alpha\beta)^2}). Since (\alpha^2+\beta^2=30) and ((\alpha\beta)^2=9), the value is (\frac{10}{3}).