Roots of a Quadratic Equation

(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta). From (52=36-2\alpha\beta), we get (\alpha\beta=-8).

Let the roots be \(t\) and \(2t\). Their sum is \(9\), so \(t+2t=9\), giving \(t=3\). Hence, the roots are \(3\) and \(6\). The product of the roots equals \(c\), so \(c=3\times6=18\). Exam tip: In \(x^2+bx+c=0\), the sum of the roots is \(-b\) and their product is \(c\).

Since both roots are negative and their ratio is \(1:2\), let them be \(-t\) and \(-2t\). Their product is \((-t)(-2t)=2t^2=18\), giving \(t=3\). Thus, the roots are \(-3\) and \(-6\), whose sum is \(-9\). By Vieta’s relation, the sum of the roots is \(-p\), so \(-p=-9\) and \(p=9\). Exam tip: for \(x^2+px+c=0\), the sum of the roots is \(-p\).

If the roots are \(\alpha\) and \(\beta\), then \(\alpha\beta=\frac{c}{a}\). Reciprocal roots have product \(1\), so \(c=a\). The condition \(b=0\) indicates additive inverse roots instead. Exam tip: use \(c/a\) for the product of roots.

The sum is (-\frac{b}{a}=\frac{3}{4}) and the product is (\frac{k}{4}). From (\frac{k}{4}=\frac{3}{4}), (k=3).

A quadratic equation has no real roots when its discriminant satisfies \(D<0\). Here, \(a=1\), \(b=-4\), and \(c=m+4\), so \(D=b^2-4ac=16-4(m+4)=-4m\). Thus, \(-4m<0\), which gives \(m>0\). If \(m=0\), then \(D=0\), giving two equal real roots; therefore, option C is incorrect. Exam tip: For questions about the nature of roots, first determine the sign of the discriminant.

The equation can be rewritten as \((x-(m+2))^2=0\). Hence its repeated root is \(x=m+2\), and the roots are equal for every real value of \(m\). Therefore, option A is correct. Option C gives only the special case \(m=-2\), not the complete set of valid values. Exam tip: in such questions, first look for a perfect square or verify that the discriminant \(D=b^2-4ac\) is zero.

By Vieta’s formulas, \(\alpha+\beta=9\) and \(\alpha\beta=18\). Therefore, \((\alpha-2)(\beta-2)=\alpha\beta-2(\alpha+\beta)+4=18-18+4=4\). Hence, option A is correct. Exam tip: 18 is only the product \(\alpha\beta\); the required expression also involves the sum of the roots.

For a quadratic equation \(ax^2+bx+c=0\), the product of its roots is \(\frac{c}{a}\). Here, \(a=5\) and \(c=q\), so the product of the roots is \(\frac{q}{5}\). Therefore, \(\frac{q}{5}=3\times(-4)=-12\). The distractor \(12\) results from missing the negative sign. Exam tip: remember that the sum of roots is \(-\frac{b}{a}\), while their product is \(\frac{c}{a}\).

The sum of the roots is \(5+(-2)=3\). Since the sum of roots is \(-p\), we get \(p=-3\). Their product is \(q=5\times(-2)=-10\). Therefore, \(p-q=-3-(-10)=7\). Exam tip: For \(x^2+px+q=0\), the sum of roots is \(-p\) and their product is \(q\).

For the quadratic equation c(x^2-6x+2=0c), the sum of the roots is c(\alpha+\beta=6c) and their product is c(\alpha\beta=2c). Using the identity c(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\betac), we get c(\alpha^2+\beta^2=6^2-2(2)=36-4=32c). Therefore, option A is correct. Exam tip: use c(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\betac); 36 is only c((\alpha+\beta)^2c), not the required value.

Here (\alpha+\beta=-2) and (\alpha\beta=-2). (\alpha^2+\alpha\beta+\beta^2=(\alpha+\beta)^2-\alpha\beta=6).

If the roots are \(\alpha\) and \(\beta\), Vieta’s formulas give \(\alpha+\beta=2m+3\) and \(\alpha\beta=(m+1)(m+2)\). For option A, \((m+1)+(m+2)=2m+3\), and the product is exactly \((m+1)(m+2)\). Hence the roots are \(m+1\) and \(m+2\). Exam tip: compare the sum and product of the proposed roots with the coefficients of the quadratic.

By Vieta’s formulas, \(\alpha+\beta=10\) and \(\alpha\beta=24\). Therefore, \((\alpha-2)(\beta-2)=\alpha\beta-2(\alpha+\beta)+4=24-20+4=8\). Hence, option A is correct. Exam tip: when subtracting the same number from both roots, expand the product carefully; using only \(\alpha\beta\) would incorrectly give 24.

The direct answer is option A: 1. To solve the equation x²−13x+42=0, look for two numbers with product 42 and sum 13. The numbers are 6 and 7, so x²−13x+42=(x−6)(x−7). Setting each factor equal to zero gives x=6 or x=7. Thus the smaller root is α=6 and the larger root is β=7. The required difference is β−α=7−6=1. Option A is correct because it states 1. Option B, 13, is the sum of the roots, 6+7=13, not their difference. Option C, 42, is the product of the roots, 6×7=42, not the difference. Option D, 6, is the smaller root itself, not the gap between the two roots. Always arrange the roots in the stated order before subtracting: larger minus smaller, not the reverse.

For a quadratic equation \(ax^2+bx+c=0\), the product of its roots is \(\alpha\beta=\frac{c}{a}\). Here, \(a=1\) and \(c=12\), so \(\alpha\beta=12\). Therefore, \(\alpha^2\beta^2=(\alpha\beta)^2=12^2=144\). Option 12 is only the value of \(\alpha\beta\), not its square. Exam tip: Remember that the sum and product of the roots are \(-\frac{b}{a}\) and \(\frac{c}{a}\), respectively.

The governing concept is the relationship between the roots and the coefficients of a quadratic equation, together with the fact that the labels α and β may be interchanged. Factoring gives x^2 - 9x + 20 = (x - 5)(x - 4), so the two roots are 5 and 4. Hence α + β = 5 + 4 = 9, which also follows from the coefficient rule α + β = -b/a = 9. If α = 5 and β = 4, then α - β = 1 and the required ratio is 9/1 = 9. If the labels are reversed, α = 4 and β = 5, then α - β = -1 and the ratio is 9/(-1) = -9. Therefore the possible values are 9 or -9, so option A is correct. Option B gives values that would require a different root difference, while options C and D do not follow from the actual sum and difference of the roots. The duplicate-option defect has been removed.

For a quadratic equation \(ax^2+bx+c=0\), the sum of the roots is \(\alpha+\beta=-\frac{b}{a}\) and their product is \(\alpha\beta=\frac{c}{a}\). Thus, here \(\alpha+\beta=4\) and \(\alpha\beta=-5\). Therefore, \((\alpha+2)(\beta+2)=\alpha\beta+2(\alpha+\beta)+4=-5+2(4)+4=7\). Hence, option A is correct. Exam tip: Use the sum and product of roots directly instead of solving for the roots individually.

Let the equal root be \(r\). Since the product of the roots is \(49\), we have \(r^2=49\), giving \(r=\pm7\). Because the roots are negative, \(r=-7\). Thus, their sum is \(-14\). By Vieta’s formula, the sum of the roots is \(-k\), so \(-k=-14\) and \(k=14\). Exam tip: use the constant term to find the equal root first, then use the sum of roots \(-k\) to determine the coefficient.

Here (\alpha+\beta=14) and (\alpha\beta=45). (\alpha^2+\beta^2=196-90=106), so the value is (\frac{106}{45}).

For \(ax^2+bx+c=0\), roots are real and equal when the discriminant \(D=b^2-4ac\) is zero. In option A, \(D=(-6)^2-4(1)(9)=36-36=0\). In option B, \(D=-4\), so its roots are not real. Exam tip: check \(D=0\) first for equal roots.

The old sum is (7) and product is (10). The reciprocal roots have sum (\frac{7}{10}) and product (\frac{1}{10}).

Here (\alpha+\beta=5) and (\alpha\beta=5). (\alpha^2+\beta^2+\alpha\beta=(\alpha+\beta)^2-\alpha\beta=20).

By Vieta’s formulas, \(\alpha+\beta=3\) and \(\alpha\beta=-28\). Therefore, \((\alpha+1)(\beta+1)=\alpha\beta+\alpha+\beta+1=-28+3+1=-24\). Hence, option A is correct. Remember to include the final \(+1\) term when expanding the product; omitting it leads to an incorrect result.

Here (\alpha+\beta=11) and (\alpha\beta=24). The value is (\frac{\alpha+\beta+2}{\alpha\beta+\alpha+\beta+1}=\frac{13}{36}).