Nature of Roots

For a quadratic equation to have real roots, its discriminant must satisfy \(D\ge0\). Here, \(a=1\), \(b=2k\), and \(c=k^2-4k+8\). Thus, \(D=b^2-4ac=4k^2-4(k^2-4k+8)=16(k-2)\). Therefore, \(16(k-2)\ge0\), giving \(k\ge2\). At \(k=2\), the roots are equal, so equality must be included. Exam tip: use \(D\ge0\) for real roots, whereas distinct real roots require \(D>0\).

For a quadratic equation to have real and equal roots, its discriminant must be zero: \(D=b^2-4ac=0\). Here, \(a=1\), \(b=2k\), and \(c=k^2-4k+8\). Thus, \(D=(2k)^2-4(k^2-4k+8)=16(k-2)\). Setting this equal to zero gives \(k=2\), so option A is correct. Exam tip: For equal roots, immediately apply the condition \(D=0\).

Here, \(a=k-2\), \(b=-2(k+1)\), and \(c=k\). For real and equal roots, the discriminant must be zero: \(D=b^2-4ac=0\). Thus, \(D=4(k+1)^2-4(k-2)k=16k+4\). Setting this equal to zero gives \(k=-\frac{1}{4}\). For this value, \(a\neq 0\), so the equation remains quadratic. Exam tip: For equal-root questions, set the discriminant to zero and then verify that the coefficient of \(x^2\) is non-zero.

A quadratic equation has no real roots exactly when its discriminant is negative. For x²+2(a−2)x+2a+5=0, the coefficients are A=1, B=2(a−2), and C=2a+5. Thus D=[2(a−2)]²−4(1)(2a+5)=4(a−2)²−8a−20=4(a²−6a−1). We require D<0, so a²−6a−1<0. The corresponding equality has solutions a=3±√10. Because the coefficient of a² is positive, the expression is negative strictly between these two values. Therefore 3−√10<a<3+√10, making option A correct. At either endpoint D=0, producing equal real roots, while outside the interval D>0, producing two distinct real roots.

For real and equal roots, the discriminant must satisfy \(D=b^2-4ac=0\). Here, \(a=r+2\), \(b=-2(r+5)\), and \(c=r+2\). Thus, \(D=4(r+5)^2-4(r+2)^2=12(2r+7)\). Setting \(D=0\) gives \(2r+7=0\), so \(r=-\frac{7}{2}\). The value \(r=-2\) is not valid because it makes the coefficient of \(x^2\) zero, so the equation is no longer quadratic. Exam tip: For equal-root questions, set \(D=0\) first and then verify that \(a\neq0\).

Here, a=1, b=−2(m−4), and c=m²−16. Therefore, the discriminant is \(D=b^2-4ac=32(4-m)\). Real and distinct roots require \(D>0\), so \(32(4-m)>0\), which gives \(m<4\). Note that when \(m=4\), \(D=0\) and the roots are equal; hence \(m\le4\) is not the correct condition. Exam tip: determine the sign of the discriminant first when analysing the nature of quadratic roots.

A quadratic equation has no real roots when its discriminant satisfies \(D<0\). Here, \(D=[2(a+3)]^2-4(a^2+10a+17)=-16(a+2)\). Therefore, \(-16(a+2)<0\), which gives \(a>-2\). At \(a=-2\), \(D=0\), so the equation has equal real roots rather than no real roots. Exam tip: simplify the discriminant completely before applying its sign condition.

For real roots, (D\ge0) is required. Here (D=4(p+2)(2p-9)), so (p\le -2) or (p\ge \frac{9}{2}).

For the equation, \(a=2\), \(b=-(q+4)\), and \(c=2q\). Equal roots require the discriminant \(D=b^2-4ac\) to be zero. Thus, \(D=(q+4)^2-16q=q^2-8q+16=(q-4)^2\). Setting this equal to zero gives \(q=4\). The values in option B do not make the discriminant zero. Exam tip: For equal roots, immediately apply the condition \(D=0\).

The governing concept is the discriminant criterion for equal roots. For a quadratic equation ax² + bx + c = 0 to have two real and equal roots, its discriminant D = b² − 4ac must be zero. Here a = k + 3, b = 2k, and c = k − 1. Therefore, D = (2k)² − 4(k+3)(k−1) = 4k² − 4(k² + 2k − 3) = 12 − 8k. Setting D equal to zero gives 12 − 8k = 0, so 8k = 12 and k = 3/2. At this value, a = 9/2, which is nonzero, so the equation remains genuinely quadratic. Hence option A is correct. The other choices do not make the discriminant zero; k = −3 would even remove the x² term.

For equal roots, a quadratic equation must satisfy the condition \(D=b^2-4ac=0\). Here, \(a=t-1\), \(b=-2(t+2)\), and \(c=t+5\). Thus, \(D=4(t+2)^2-4(t-1)(t+5)=4[(t+2)^2-(t^2+4t-5)]=36\), which is never zero for any real \(t\). Therefore, no real value of \(t\) produces equal roots. Also, when \(t=1\), the coefficient of \(x^2\) becomes zero, so the equation is linear rather than quadratic. Exam tip: check \(a\neq0\) first, and then apply the equal-roots condition \(D=0\).

For real roots, the discriminant must satisfy \(D\ge0\). Here \(a=1\), \(b=-2(c+1)\), and the constant term is \(c^2-2c+5\), so \(D=b^2-4ac=16(c-1)\). Thus \(16(c-1)\ge0\), giving \(c\ge1\). At \(c=1\), the equation has equal real roots, so the stricter condition \(c>1\) is incorrect. Exam tip: For a quadratic to have real roots, begin by imposing \(D\ge0\).

Here, \(a=3\), \(b=-2\sqrt{15}\), and \(c=5\). The discriminant is \(D=b^2-4ac=(-2\sqrt{15})^2-4(3)(5)=60-60=0\). When \(D=0\), the two roots are real and equal. In fact, the repeated root is \(x=\frac{-b}{2a}=\frac{\sqrt{15}}{3}\). Remember that the roots may be irrational and still equal; therefore, option D is incorrect because it states that the roots are distinct.

The direct answer is A: the roots are real, irrational, and distinct. Here a=1, b=-2√7, and c=3. The discriminant is D=b^2-4ac=(-2√7)^2-4(1)(3)=28-12=16, which is positive, so the roots are real and different. Using the quadratic formula, x=(2√7 plus or minus 4)/2=√7 plus or minus 2. Since √7 is irrational, adding or subtracting the rational number 2 still gives irrational numbers. Option A is correct. Option B is wrong because D is not zero. Option C is wrong because D is positive. Option D is wrong because the roots are not rational. The key test is: positive non-square-related root expressions may be irrational; here the explicit roots show this clearly.

For a quadratic equation to have real and equal roots, its discriminant must be zero. Here, \(a=1\), \(b=-2\sqrt{n}\), and \(c=4\), so \(D=b^2-4ac=(-2\sqrt{n})^2-4(1)(4)=4n-16\). Setting \(D=0\) gives \(4n-16=0\), hence \(n=4\). As an exam tip, use \(D=0\) immediately whenever the roots are stated to be equal and real.

Here (D=4[(k+1)^2-5(k-3)]=4(k^2-3k+16)). This expression is always positive, so real roots always exist.

For x² + 2(2y−1)x + 4y² + 3 = 0, the coefficients are a = 1, b = 2(2y−1) and c = 4y² + 3. The discriminant is D = [2(2y−1)]² − 4(4y²+3) = 4(2y−1)² − 16y² − 12. Expanding gives 4(4y²−4y+1) − 16y² − 12 = −16y − 8 = −8(2y+1). Real roots require D ≥ 0, so −8(2y+1) ≥ 0, which gives y ≤ −1/2. Equality at y = −1/2 gives equal roots; smaller values give distinct real roots. Therefore option A is correct. The other statements incorrectly claim an unconditional result or restrict y to an incorrect single value.

Here (D=4(3u+1)^2-4(9u^2-6u+5)=48u-16). Thus the nature depends on (u), not always distinct.

For a quadratic equation to have real and distinct roots, its discriminant must satisfy \(D>0\). Thus, \((s-2)(s+5)>0\). The zero points are \(s=2\) and \(s=-5\); the product is positive outside these points, giving \(s<-5\) or \(s>2\). In option B, \(D<0\), so the roots are non-real, while in option C, \(D=0\), so the roots are equal. Exam tip: for a product inequality, mark the zero points on a number line and check the sign in each interval.

A quadratic equation has no real roots when its discriminant satisfies \(D<0\). Thus, \((z+1)^2-9<0\), which gives \((z+1)^2<9\). Therefore, \(-3<z+1<3\), and hence \(-4<z<2\). So, option A is correct. Remember that \(D=0\) gives two equal real roots, so the endpoints \(-4\) and \(2\) are excluded from the interval.

A quadratic equation has equal roots when its discriminant is zero. Here, \(D=[-(a+6)]^2-4(1)(6a)=(a+6)^2-24a=(a-6)^2\). Thus, \((a-6)^2=0\), giving \(a=6\). Exam tip: For equal-root questions, set the discriminant directly equal to zero; the other listed values produce distinct roots.

Here, \(A=1\), \(B=-2(a+b)\), and \(C=4ab\). Therefore, the discriminant is \(D=B^2-4AC=4(a+b)^2-16ab=4(a-b)^2\). Since \(a\ne b\), we have \((a-b)^2>0\), so \(D>0\). Hence, the roots are real and distinct. Equal roots occur only when \(D=0\), so option B is not correct. Exam tip: for a quadratic equation, \(D>0\) indicates two real and distinct roots.

A quadratic equation has equal real roots when its discriminant is zero. Here, \(D=(m+n)^2-4mn=m^2-2mn+n^2=(m-n)^2\). Thus \((m-n)^2=0\), which gives \(m=n\). Exam tip: For equal-root questions, begin by setting the discriminant equal to zero.

For the quadratic equation, the discriminant is \(D=[2(a-b)]^2-4(a+b)^2\). Hence, \(D=-16ab\). Real roots require \(D\ge 0\), so \(-16ab\ge 0\), which gives \(ab\le 0\). Option D is also incorrect because equal roots require \(D=0\), giving \(ab=0\), not generally \(a=b\). Exam tip: To determine whether quadratic roots are real, first check the sign of the discriminant.

The direct answer is A: the roots are always real and distinct, assuming ℓ is real. For the quadratic x^2−2ℓx+ℓ^2−9=0, the coefficients are a=1, b=−2ℓ, and c=ℓ^2−9. The discriminant is D=b^2−4ac = (−2ℓ)^2−4(1)(ℓ^2−9) = 4ℓ^2−4ℓ^2+36 = 36. Since 36 is positive for every real value of ℓ, the equation always has two real and unequal roots. In fact, rewriting gives (x−ℓ)^2=9, so x=ℓ+3 or x=ℓ−3; these differ by 6. Option A is correct. Option B is wrong because equal roots require D=0, not D=36. Option C is wrong because non-real roots require D<0. Option D is wrong because the roots are real for every real ℓ, not only when ℓ=3; ℓ=3 is not a special requirement. The condition that ℓ be real is understood in this school context. Memory cue: positive discriminant means real and distinct roots.