Roots of a Quadratic Equation

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 roots are (6) and (7). Thus the smaller root is (6) and the larger root is (7), so (\beta-\alpha=1).

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.

Factorising x^2-9x+20 gives (x-5)(x-4)=0, so the roots are 5 and 4. Their sum is α+β=9. Since the labels α and β are interchangeable, α-β may be 5-4=1 or 4-5=-1. Therefore the ratio is 9/1=9 or 9/(-1)=-9. Hence option A is the intended correct answer. The displayed option B duplicates option A exactly, which is a content-quality defect because the alternatives are not distinct; it should be replaced by a different value, such as 9/3 or -9/3, while preserving A as the unique correct choice. Options C and D do not result from the root sum and difference.

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}).

For reciprocal roots, the product is (1), and here the product is (k). Hence (k=1); in exams, check the product first.

Here (\alpha+\beta=\frac{5}{2}) and (\alpha\beta=\frac{m}{2}). Using (\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta), we get (m=3).

For a quadratic equation \(ax^2+bx+c=0\) to have equal roots, its discriminant must be zero: \(D=b^2-4ac=0\). Here, \(a=k\), \(b=-2(k+1)\), and \(c=k+4\). Therefore, \(D=4(k+1)^2-4k(k+4)=4(1-2k)\). Setting this equal to zero gives \(k=\frac{1}{2}\). Exam tip: In equal-root questions, begin with \(D=0\) and verify that \(a\neq 0\) so the equation remains quadratic.

For a quadratic equation to have real roots, its discriminant must satisfy \(D\ge 0\). Here, \(D=[-2(a+1)]^2-4(a^2+3)=8(a-1)\). Therefore, \(8(a-1)\ge 0\), which gives \(a\ge 1\). Option A incorrectly excludes \(a=1\), although at \(a=1\) the equation has equal real roots. Exam tip: For questions about real roots, begin by applying \(D\ge 0\).

Let the positive integer roots be r and s, with r>s. Vieta’s product relation gives rs=36, and the stated difference gives r-s=5. The positive factor pairs of 36 are (1,36), (2,18), (3,12), and (4,9). Only 9-4=5, so the roots must be 9 and 4. Their sum is 9+4=13. For x^2-px+36=0, Vieta’s sum relation says r+s=p, because the coefficient of x is -p. Therefore p=13, making option C correct. The other options do not equal the sum of the only positive factor pair whose difference is 5. This also confirms that the quadratic is x^2-13x+36=(x-9)(x-4).