Roots of a Quadratic Equation

The governing concept is factorisation of a quadratic equation. For x^2+3x−18, we need two numbers whose product is −18 and whose sum is 3, because the coefficient of x is 3. The numbers 6 and −3 satisfy these conditions, so x^2+3x−18=(x+6)(x−3). Therefore, the roots are x=3 and x=−6, making option A correct. Option B has the correct numbers with reversed signs, but their sum is 3 and product is −18 only for 3 and −6, not −3 and 6. The other pairs do not have the required sum and product.

For equal roots, (s^2-36=0), so (s=\pm6). The equal root is (\frac{s}{2}), which is positive when (s=6).

For a quadratic equation (ax^2+bx+c=0) to have equal roots, its discriminant (D=b^2-4ac) must be zero. Here, a=k-2, b=4, c=1, so D=16-4(k-2)=0, giving k=6. Note that when k=2, the equation becomes linear rather than quadratic, so it cannot represent equal roots of a quadratic equation. Exam tip: Set the discriminant equal to zero whenever a quadratic is stated to have equal roots.

((\alpha-2)(\beta-2)=\alpha\beta-2(\alpha+\beta)+4). Since (\alpha+\beta=6) and (\alpha\beta=8), the value is (0).

Here (\alpha+\beta=\frac{11}{3}) and ((\alpha-\beta)^2=\frac{25}{9}). Using ((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta), we get (\alpha\beta=\frac{8}{3}), so (p=8).

For equal roots, put (D=0). Simplifying (4(k+2)^2-4(k^2+5)=0) gives (4k-1=0), so (k=\frac{1}{4}).

The governing idea is transformation of roots. From x^2−6x+5=(x−1)(x−5), the original roots are α=1 and β=5. Applying the stated transformation gives new roots 3(1)−2=1 and 3(5)−2=13. A monic quadratic with roots 1 and 13 is x^2−(1+13)x+(1)(13)=0, hence x^2−14x+13=0. Therefore option A is correct. Option B has the wrong sum and product, option C keeps the same sum but has an incorrect product, and option D uses the wrong sign for the sum of the roots.

Let the roots be (r) and (2r), then (2r^2=4) gives (r=\pm\sqrt{2}). Since (3r=-\frac{m}{2}), we get (m=\pm6\sqrt{2}).

The governing concept is verification of a proposed root by substitution, followed by factorisation. Substitute x=4 into x^2−(a+4)x+4a: 16−4(a+4)+4a=16−4a−16+4a=0 for every value of a. Thus 4 is always a root. In fact, the equation factors as (x−4)(x−a)=0, so the other root is a. Option B is not always correct because a+4 is generally not a root. The roots are equal only when a=4, and they are real for all real a, so options C and D are false.

We use (\frac{1}{\alpha^2}+\frac{1}{\beta^2}=\frac{\alpha^2+\beta^2}{(\alpha\beta)^2}). Since (\alpha^2+\beta^2=21) and ((\alpha\beta)^2=4), the value is (\frac{21}{4}).

For reciprocal roots, (\frac{k}{k}=1), so (k\neq0) is needed. For real roots, (64-4k^2\ge0), hence (k^2\le16).

By Vieta’s formulas, the product of the roots is 12, so (r(r+1)=12), or (r^2+r-12=0). Thus, r=3 or r=-4. The sum of the roots is therefore 7 or -7. Since r+(r+1)=-p, we obtain p=-7 or p=7; hence, option A is correct. Exam tip: for x^2+px+q=0, the sum of the roots is -p and their product is q.

A quadratic equation has real roots only when its discriminant satisfies \(D\ge0\). Here, \(A=4\), \(B=-4(a-1)\), and \(C=a^2-4a\). Thus, \(D=B^2-4AC=16(a-1)^2-16(a^2-4a)=16(2a+1)\). Therefore, \(16(2a+1)\ge0\), giving \(a\ge-\frac12\). Hence, option A is correct; the condition \(a\le1\) in option B does not follow from the discriminant. Exam tip: simplify the discriminant completely before solving the parameter inequality.

The roots are (3) and (5). Substitution gives (\frac{5}{1}+\frac{7}{3}=\frac{22}{3}), so none of the options is correct; the correct value should be (\frac{22}{3}).

One root can be (\alpha=2), which makes (\alpha-2=0). Therefore the expression is undefined; always check zero denominators first in exams.

Here (\alpha+\beta=\frac{5}{2}) and (\alpha\beta=\frac{3}{2}). Using (\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)), we get (\frac{35}{8}).

For equal roots, the discriminant must be zero. Here, \(a=1\), \(b=m-5\), and \(c=9\), so \(D=b^2-4ac=(m-5)^2-36=0\). Hence, \(m-5=\pm6\), giving \(m=11\) or \(m=-1\). Also, the product of the roots is \(c/a=9>0\), so the roots cannot have opposite signs. Exam tip: For equal roots of a quadratic equation, directly use \(b^2-4ac=0\).

For a quadratic equation \(x^2+ax+b=0\), the sum of the roots is \(-a\) and their product is \(b\). Here, the sum is \((2+\sqrt{3})+(2-\sqrt{3})=4\), so \(a=-4\). Their product is \((2+\sqrt{3})(2-\sqrt{3})=4-3=1\), so \(b=1\). Therefore, \(a+b=-4+1=-3\). Exam tip: Use Vieta’s formulas directly to find the sum and product of the roots.

Since (\alpha^2=3\alpha+1), the expression becomes (3\alpha+1+3\beta-1=3(\alpha+\beta)=9), so no option is correct. The correct value is (9).

Use (\alpha^2=3\alpha+1) and (\alpha\beta=-1). Then the expression is (3\alpha+1+3\beta-1=3(\alpha+\beta)=9).