Roots of a Quadratic Equation

By Vieta’s formulas, the sum of the roots is \(-p\) and their product is \(q\). Here, the sum is \((-2)+(-5)=-7\), so \(-p=-7\) gives \(p=7\). Their product is \((-2)(-5)=10\), hence \(q=10\). Therefore, \(p+q=7+10=17\). In exams, match the sum and product of the roots with \(-p\) and \(q\), respectively.

For a quadratic equation \(x^2+bx+c=0\), the product of the roots is \(\alpha\beta=c\). Here, \(c=k\), so \(k=(-2)(-4)=8\). The option \(-8\) results from an incorrect sign in multiplying two negative numbers. Exam tip: Verify the roots using both their sum \((-b)\) and product \(c\).

The sum of reciprocals is (\frac{\alpha+\beta}{\alpha\beta}). Here (\frac{\frac{7}{2}}{\frac{3}{2}}=\frac{7}{3}).

(\alpha+\beta=9) and (\alpha\beta=20). (\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)=729-540=189).

For real and equal roots, the discriminant must be zero. Here, \(a=1, b=-4, c=k\), so \(D=b^2-4ac=16-4k\). Thus, \(16-4k=0\), giving \(k=4\). The value 2 may result from an incorrect calculation of the discriminant equation. Exam tip: Set \(D=0\) whenever a quadratic equation has equal roots.

For equal roots, the discriminant \(D=b^2-4ac\) must be zero. Here, \(a=k\), \(b=-6\), and \(c=9\), so \(D=(-6)^2-4(k)(9)=36-36k\). Setting this equal to zero gives \(k=1\). Exam tip: equal roots in a quadratic equation always require \(D=0\).

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(D=2^2-4(1)(5)=4-20=-16<0\), so the equation has no real roots. Therefore, the correct answer is 0. One or two real roots are possible only when the discriminant is 0 or positive, respectively. Exam tip: Check the sign of the discriminant first.

For a monic quadratic x² − Sx + P = 0, the sum of the roots is S and their product is P. Taking the proposed roots as 1 and a, their sum is 1 + a = a + 1, while their product is 1 × a = a. These values agree exactly with the coefficient pattern x² − (a + 1)x + a = 0. Therefore option A correctly explains the statement. Option B interchanges the sum and product. The roots are equal only for the special value a = 1, not for every a, so option C is not generally valid. The discriminant is (a − 1)², which is never negative, contradicting option D.

Here (\alpha+\beta=6) and (\alpha\beta=8). ((\alpha+1)(\beta+1)=\alpha\beta+\alpha+\beta+1=15).

(\alpha^2\beta+\alpha\beta^2=\alpha\beta(\alpha+\beta)). Here (\alpha\beta=-10) and (\alpha+\beta=3), so the value is (-30).

Here (\alpha+\beta=\frac{5}{2}) and (\alpha\beta=1). (\frac{1}{\alpha^2}+\frac{1}{\beta^2}=\frac{(\alpha+\beta)^2-2\alpha\beta}{(\alpha\beta)^2}=\frac{17}{4}).

For equal roots, the roots are (-4) and (-4) because the product is (16). The sum is (-8), so (-p=-8) gives (p=8).

Putting (x=1) gives (1+2a=0), so (a=-\frac{1}{2}). The product is (a), so the other root is (-\frac{1}{2}).

For x² − 5x + q = 0, Vieta’s relations give the sum of the roots as 5. Let the roots be r and s, with r − s = 1. Solving r + s = 5 and r − s = 1 gives r = 3 and s = 2. Their product is q, so q = rs = 3 × 2 = 6. Therefore option A is correct. The other choices do not satisfy both the required sum and difference.

Here the sum of roots is (S) and product is (P). Therefore (\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=S^2-2P).

The old roots are (6) and (-2). The new roots are (7) and (-1), so the equation is (x^2-6x-7=0).

Factoring the equation as \((x-6)(x+2)=0\) gives the roots 6 and -2. After subtracting 1 from each root, the new roots are 5 and -3, so their product is \(5\times(-3)=-15\). In such questions, finding the original roots first is a direct and reliable method.

For the original equation x² − 3x + 2 = 0, Vieta’s relations give α + β = 3 and αβ = 2. When each root is multiplied by 2, the new roots are 2α and 2β. Their sum becomes 2α + 2β = 2(α + β) = 6, and their product becomes (2α)(2β) = 4αβ = 8. A monic quadratic with these new roots is x² − 6x + 8 = 0. Thus option A is correct. Option C changes the sum correctly but fails to multiply the product by 4; option B leaves the sum unchanged, and option D uses the wrong sign.

By Vieta’s formula, \(\alpha+\beta=\frac{-(-5)}{1}=5\). Therefore, \((\alpha+2)+(eta+2)=\alpha+\beta+4=5+4=9\). Hence, the correct answer is 9. Since 2 is added to both roots, the sum increases by 4, not by 2; therefore, 7 is incorrect. Exam tip: when the same number \(k\) is added to both roots, their sum increases by \(2k\).

The given equation can be written as \(x^2-10x+25=(x-5)^2=0\). Thus, both roots are equal: \(\alpha=\beta=5\). Therefore, \(\alpha-\beta=5-5=0\), so option A is correct. Exam tip: When the discriminant of a quadratic equation is zero, its two roots are equal.