Roots of a Quadratic Equation

Here (\alpha+\beta=12) and (\alpha\beta=32). (\alpha^2+\beta^2=144-64=80), and (\frac{80}{32}=\frac{5}{2}).

The sum of the roots is \((-1)+(-4)=-5\). For a quadratic equation \(ax^2+bx+c=0\), the sum of the roots is \(-\frac{b}{a}\). Therefore, \(-\frac{p}{2}=-5\), giving \(p=10\). Option C, \(8\), is the constant term, not the value of \(p\). Exam tip: Check the sum and product of roots using \(-\frac{b}{a}\) and \(\frac{c}{a}\), respectively.

The old sum is (5) and product is (6). The reciprocal roots have sum (\frac{5}{6}) and product (\frac{1}{6}).

Here (\alpha+\beta=8) and (\alpha\beta=15). The value is (\frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}=\frac{34}{15}).

((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta=81-80=1). This identity is quick for difference questions.

The equation can be rewritten as ((x-k)^2-4=0). Thus, ((x-k)^2=4), giving the roots (x=k+2) and (x=k-2). Therefore, their positive difference is ((k+2)-(k-2)=4). Option B, 2, is the distance of each root from k, not the difference between the two roots. Exam tip: If ((x-a)^2=b^2), the roots are (a+b) and (a-b).

For a quadratic equation \(Ax^2+Bx+C=0\), the product of its roots is \(\frac{C}{A}\). Here, the product is \(3\times5=15\), while \(\frac{C}{A}=\frac{15}{a}\). Thus, \(\frac{15}{a}=15\), giving \(a=1\). The value 15 is the constant term, not the coefficient \(a\). Exam tip: use \(\frac{c}{a}\) for the product of roots and \(-\frac{b}{a}\) for their sum.

For a quadratic equation
x^2+bx+c=0
, the sum of the roots is
-b
. Here,
b=-(m+3)
, so the sum of the roots is
m+3
. Since the roots are 3 and 6,
3+6=9=m+3
, giving
m=6
. Option 3 results from mistakenly treating
m+3
as
m
. Exam tip: Verify the result using both the sum of roots and their product;
3\times6=18
matches the constant term.

By Vieta’s formulas, the sum of the roots is \(-p\) and their product is \(q\). Thus, \((-3)+(-4)=-7=-p\), giving \(p=7\). Also, \(q=(-3)(-4)=12\). Therefore, \(p+q=7+12=19\). Option 12 is only the value of \(q\). Exam tip: For \(x^2+px+q\), the sum of roots is \(-p\), while their product is \(q\).

The sum of reciprocals is (\frac{\alpha+\beta}{\alpha\beta}). Here (\frac{\frac{10}{3}}{1}=\frac{10}{3}).

From the coefficients, \(\alpha+\beta=10\) and \(\alpha\beta=21\). Using the identity \(\alpha^3+\beta^3=(\alpha+eta)^3-3\alpha\beta(\alpha+eta)\), we get \(10^3-3\times21\times10=1000-630=370\). Hence, the correct answer is 370. Exam tip: For the sum of cubes of roots, first find their sum and product using Vieta’s relations.

For real and equal roots, the discriminant must be zero. Here, \(a=1, b=-6, c=k\), so \(D=b^2-4ac=(-6)^2-4(1)(k)=36-4k\). Setting \(D=0\) gives \(36-4k=0\), hence \(k=9\). The value 6 is only the magnitude of the coefficient of \(x\), so it is not correct. Exam tip: For equal roots of a quadratic equation, always use \(b^2-4ac=0\).

For equal roots of a quadratic equation \(ax^2+bx+c=0\), the discriminant must be zero: \(D=b^2-4ac=0\). Here, \(a=k\), \(b=-8\), and \(c=16\), so \(D=(-8)^2-4(k)(16)=64-64k\). Thus, \(64-64k=0\), giving \(k=1\). Exam tip: Whenever equal roots are mentioned, immediately use the condition \(D=0\).

Here, \(a=1\), \(b=4\), and \(c=13\). The discriminant is \(D=b^2-4ac=4^2-4(1)(13)=16-52=-36<0\). Therefore, the quadratic equation has no real roots, so the answer is 0. Exam tip: \(D<0\) means zero real roots; one or two real roots occur only when \(D=0\) or \(D>0\), respectively.

By Vieta’s formulas, \(\alpha+\beta=\frac{8}{1}=8\) and \(\alpha\beta=\frac{12}{1}=12\). Therefore, \((\alpha+2)(\beta+2)=\alpha\beta+2(\alpha+\beta)+4=12+2(8)+4=32\). Hence, the correct answer is 32. In the exam, remember to include the middle term \(2(\alpha+\beta)\) when expanding the product.

(\alpha^2\beta+\alpha\beta^2=\alpha\beta(\alpha+\beta)). Here (\alpha\beta=-21) and (\alpha+\beta=4), so the value is (-84).

Here (\alpha+\beta=\frac{8}{3}) and (\alpha\beta=\frac{4}{3}). The value is (\frac{(\alpha+\beta)^2-2\alpha\beta}{(\alpha\beta)^2}=\frac{7}{4}).

Putting (x=2) gives (4+3a=0), so (a=-\frac{4}{3}). The product is (a), so the other root is (-\frac{2}{3}).

For the quadratic equation x² - 7x + q = 0, let the roots be α and β. By the relationships between roots and coefficients, α + β = 7 and αβ = q. The difference is given as 1, so take α - β = 1; if the order is reversed, the absolute difference is still 1. Solving α + β = 7 and α - β = 1 gives 2α = 8, hence α = 4, and β = 3. Therefore q = αβ = 4 × 3 = 12. Equivalently, the discriminant is (α - β)² = 1, so 49 - 4q = 1, which also gives q = 12. The other options do not satisfy both the sum and the stated root difference.

Here the sum of roots is (S) and product is (P). Therefore ((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta=S^2-4P).

The given equation factors as \(x^2-6x-16=(x-8)(x+2)\), so its roots are 8 and −2. Adding 1 to each root gives 9 and −1. Their sum is 8 and their product is −9; therefore, the required monic equation is \(x^2-8x-9=0\). Option B retains the old roots, while option C uses an incorrect sum of the new roots. Exam tip: for roots \(\alpha\) and \(\beta\), form the monic equation as \(x^2-(\alpha+\beta)x+\alpha\beta=0\).

The roots of the equation are 8 and -2, since \(x^2-6x-16=(x-8)(x+2)\). After subtracting 2 from each root, the new roots are 6 and -4, whose product is \(6\times(-4)=-24\). Remember that the transformation must be applied to both roots; changing only one root gives an incorrect result.

For the original equation, Vieta’s relations give α+β=4 and αβ=3. If the new roots are 3α and 3β, their sum is 3α+3β=3(α+β)=12, while their product is (3α)(3β)=9αβ=27. A monic quadratic with roots r and s is x^2-(r+s)x+rs=0. Substituting the new sum and product gives x^2-12x+27=0, so option A is correct. Option B keeps the old sum, option C uses the old product multiplied by only 3 instead of 9, and option D has the wrong sign for the x-term. The result can also be checked from the original roots 1 and 3, which become 3 and 9.

By Vieta’s formula, the sum of the roots of 8x^2-6x+5=09 is 8\alpha+\beta9 = -\frac{-6}{1}=6. Hence, 8\alpha+39+8\beta+39 = \alpha+\beta+6 = 6+6=12. The distractor 9 results from adding 3 only once. Exam tip: when the same number is added to both roots, their sum increases by twice that number.

The quadratic equation (x^2-12x+36=0) factors as ((x-6)^2=0). Hence both roots are 6, so (\alpha=\beta=6) and (\alpha-\beta=6-6=0). Exam tip: when the discriminant (D=b^2-4ac) is zero, the two roots are equal and their difference is zero.