Roots of a Quadratic Equation

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 \(D\ge 0\).

Here (\alpha+\beta=\frac{10}{3}) and (\alpha\beta=1). The reciprocal roots also have sum (\frac{10}{3}) and product (1).

Taking the roots as (r) and (2r), we get (3r=3-k) and (2r^2=k). Solving (2k^2-21k+18=0) gives the two listed values.

Use ((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta). Here (\alpha+\beta=\frac{7}{2}) and (\alpha\beta=\frac{3}{2}), so the value is (\frac{25}{4}).

For equal roots, the discriminant must be zero. Here, \(a=1\), \(b=2(k-1)\), and \(c=k+5\). Thus, \(D=[2(k-1)]^2-4(1)(k+5)=4(k^2-3k-4)=0\). Factoring gives \((k-4)(k+1)=0\), so \(k=4\) or \(k=-1\). The distractors arise from sign or factorisation errors. Exam tip: whenever a quadratic has equal roots, immediately apply \(D=0\).

Here (\alpha+\beta=6) and (\alpha\beta=m). Using (\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)), we get (m=8).

Put ((\alpha-\beta)^2=9) in ((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta). This gives (a^2-8a-5=0), so (a=4\pm\sqrt{21}).

The governing concept is the relation between roots and coefficients, together with transformation of roots. The original equation factors as (x−2)(x−3)=0, so α and β are 2 and 3 in either order. Adding 1 to each root produces the new roots 3 and 4. For roots r and s, the monic quadratic is x^2−(r+s)x+rs=0. Here their sum is 3+4=7 and their product is 3×4=12, giving x^2−7x+12=0. The same result follows from Vieta: α+β=5 and αβ=6, so the transformed sum is α+β+2=7 and product is (α+1)(β+1)=6+5+1=12. Therefore A is correct. B, C and D have an incorrect sum, product, or sign.

Let the roots be (r) and (4r). Then (4r^2=4), so (r=\pm1); using (5r=-\frac{p}{3}), we get (p=\pm15).

If the roots are \(r\) and \(-r\), their sum is 0; by Vieta, \(-b/a=0\), so \(b=0\). For distinct real roots, \(ac<0\). With \(ac>0\), the roots are non-real. Exam tip: check both sum and product.

For real roots, (D=36-36k\ge0) is required. Thus (k\le1), and (k\ne0) is also needed for a quadratic equation.

Let the roots be \(\alpha\) and \(-\alpha\). Their sum is \(\alpha+(-\alpha)=0\). By Vieta’s formula, the sum of the roots of \(x^2+bx+c=0\) is \(-b\). Hence \(-b=0\), so \(b=0\). The condition \(c=0\) is not necessary; for example, \(x^2-1=0\) has roots \(1\) and \(-1\), but \(c=-1\). Exam tip: compare the sum of the roots directly with the coefficient relation \(-b\).

(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}). With (\alpha+\beta=-4) and (\alpha\beta=1), the value is (14).

Here, \(a=5, b=-2, c=1\). The discriminant is \(D=b^2-4ac=(-2)^2-4(5)(1)=4-20=-16<0\). Therefore, the equation has no real roots. Option A would be correct only if \(D=0\), while option B requires \(D>0\). Exam tip: The sign of the discriminant quickly determines the nature of the roots of a quadratic equation.

A root must satisfy the equation. Substituting \(x=2\), we get \(k(2)^2-6(2)+4=0\), so \(4k-12+4=0\). Hence \(4k=8\) and \(k=2\). Therefore, option B is correct. Exam tip: Substitute the given root directly into the quadratic equation to find the unknown coefficient.

For a quadratic equation \\(Ax^2+Bx+C=0\\), the sum of the roots is \\(-B/A\\), provided \\(A\\ne0\\). Here, \\(A=m+1\\) and \\(B=-2(m-1)\\), so the sum is \\(2(m-1)/(m+1)\\). To make this sum equal to 2, solve \\(2(m-1)/(m+1)=2\\), with the necessary restriction \\(m\\ne-1\\).

Dividing by 2 and multiplying by \\(m+1\\) gives \\(m-1=m+1\\). Subtracting \\(m\\) from both sides gives \\(-1=1\\), which is impossible. The excluded value \\(m=-1\\) makes the quadratic coefficient zero, so it cannot provide a quadratic equation anyway. Therefore no permitted value of \\(m\\) works, and option D is correct.

Here (\alpha+\beta=3) and (\alpha\beta=-2). Thus (\alpha^2+\beta^2=13) and (\alpha^2\beta^2=4), so the equation is (x^2-13x+4=0).

The sum of roots is (4) and the product is (-1). Use (x^2-(\text{sum})x+\text{product}=0) to get the answer.

For reciprocal roots, (\alpha\beta=1). Here (\alpha\beta=k), so (k=1), and (D=12>0) confirms real roots.

For equal roots, the discriminant \(D=b^2-4ac\) must be zero. Here, \(a=2\), \(b=\lambda\), and \(c=8\), so \(D=\lambda^2-4(2)(8)=\lambda^2-64\). Thus, \(\lambda^2-64=0\), giving \(\lambda=\pm8\). Option B may result from incorrectly calculating \(4ac\). Exam tip: For equal roots of a quadratic equation, set the discriminant directly equal to zero.

The denominator ((\alpha-1)(\beta-1)=\alpha\beta-(\alpha+\beta)+1=4). The numerator is (\alpha+\beta-2=5), so the value is (\frac{5}{4}).

For roots \(\alpha\) and \(\beta\) of \(ax^2+bx+c=0\), \((\alpha-\beta)^2=\frac{b^2-4ac}{a^2}\). Here, \(a=1, b=-(k+2), c=2k\), so \((\alpha-\beta)^2=(k+2)^2-8k=(k-2)^2\). Since the difference is 2, \((k-2)^2=4\), giving \(k-2=\pm2\), hence \(k=0\) or \(k=4\). Therefore, option A is correct. Exam tip: Relate the square of the difference between the roots to the discriminant before solving for the parameter.