Roots of a Quadratic Equation

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.

The key idea is testing a constant candidate root and identifying the factorisation. Substituting x=6 gives 36−6(a+6)+6a=36−6a−36+6a=0, which is true for every value of a. Thus 6 is always a root. The complete factorisation is (x−6)(x−a)=0, so the second root is a. Option A is not generally true: substituting x=a+6 does not give zero for arbitrary a. The roots are equal only when a=6, and for real a both roots are real, so options C and D cannot be correct.

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}).

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).