Nature of Roots

Equal roots require the discriminant to be zero. Here \(a=2\), \(b=-(k+4)\), and \(c=2k\), so \(D=b^2-4ac=(k+4)^2-16k=(k-4)^2\). Thus, \((k-4)^2=0\) gives \(k=4\). For example, when \(k=-4\), \(D=64\), so the roots are not equal. Exam tip: For equal roots of a quadratic equation, set \(b^2-4ac=0\).

Here (D=4(k-1)^2-4(k+5)=4(k^2-3k-4)). Use (D<0) and factor carefully before choosing the interval.

A quadratic equation has no real roots when its discriminant satisfies \(D<0\). Here, \(a=1\), \(b=2(k-1)\), and \(c=k+5\). Hence, \(D=[2(k-1)]^2-4(k+5)=4(k^2-3k-4)=4(k-4)(k+1)\). Therefore, \((k-4)(k+1)<0\), which holds for \(-1<k<4\). At \(k=-1\) or \(k=4\), \(D=0\), so the roots are equal and real, not absent. Exam tip: for questions on the nature of roots, first calculate \(D\) and then analyse its sign.

For equal roots, the discriminant \(D=b^2-4ac\) must be zero. Here, \(a=k\), \(b=-2(k+1)\), and \(c=k+3\). Therefore, \(D=4(k+1)^2-4k(k+3)=4(1-k)\). Setting \(D=0\) gives \(k=1\), which also satisfies \(k\neq0\). Exam tip: For equal roots of a quadratic equation, apply \(D=0\) directly.

Here (D=(2k)^2-4(k-2)(k+3)=4(6-k)). For real roots we need (k\leq6), so check simplification carefully.

For a quadratic equation \\(Ax^2+Bx+C=0\\), real roots exist when the discriminant \\(D=B^2-4AC\\) is non-negative. Here, \\(A=k-2\\), \\(B=2k\\), and \\(C=k+3\\). Therefore, \\(D=(2k)^2-4(k-2)(k+3)=4k^2-4(k^2+k-6)=4(6-k)\\). For real roots, \\(4(6-k)\\ge0\\), which gives \\(k\\le6\\).

The question separately states that \\(k\\ne2\\), because at \\(k=2\\) the coefficient of \\(x^2\\) becomes zero and the equation is no longer quadratic. Combining both conditions gives \\(k\\le6\\) and \\(k\\ne2\\). Hence option A is correct. The condition \\(D=0\\) is included because equal real roots are still real roots.

A quadratic equation has real and distinct roots only when its discriminant is positive. Here, \(A=1\), \(B=-2(a+b)\), and \(C=(a-b)^2\). Thus, \(D=[-2(a+b)]^2-4(a-b)^2=16ab\). The condition \(D>0\) gives \(16ab>0\), and hence \(ab>0\). If \(ab=0\), the roots are equal, while \(ab<0\) gives non-real roots. Exam tip: simplify the discriminant completely before applying its sign condition.

The discriminant is \(D=[-2(a+b)]^2-4(a^2+b^2)=4(a+b)^2-4(a^2+b^2)=8ab\). For real roots, \(D\geq0\), so \(8ab\geq0\), which gives \(ab\geq0\). Option B generally gives \(D\leq0\) and does not ensure real roots. Exam tip: For a quadratic equation, first apply the condition \(D\geq0\) to test for real roots.

For \(ax^2+bx+c=0\), roots are real and distinct when the discriminant \(D=b^2-4ac>0\). In option C, \(D=(-5)^2-4(1)(6)=1>0\). Options A and D have \(D=0\), so they give equal roots. Exam tip: check the discriminant first to identify root nature.

Here, \(a=3\), \(b=-2\sqrt{6}\), and \(c=2\). Therefore, the discriminant is \(D=b^2-4ac=(-2\sqrt{6})^2-4(3)(2)=24-24=0\). When \(D=0\), the quadratic equation has two real and equal roots. Hence, option A is correct. Exam tip: \(D<0\) indicates no real roots, while \(D=0\) specifically indicates equal real roots.

For a quadratic equation ax²+bx+c=0, the discriminant D=b²−4ac determines the nature of its roots. Here a=2, b=−5√2, and c=12. Thus b²=(−5√2)²=25×2=50, while 4ac=4×2×12=96. Therefore D=50−96=−46. Since the discriminant is negative, the equation has no real roots; its two roots are complex conjugates. Hence option A is correct. D=0 would indicate equal real roots, a positive discriminant would indicate two distinct real roots, and the values D=50 or D=2 are simply incorrect calculations. The irrational coefficient does not change the discriminant rule.

Here (D=4\lambda^2-4\lambda=4\lambda(\lambda-1)). For distinct real roots (D>0), so (\lambda<0) or (\lambda>1).

A quadratic equation has equal roots when its discriminant is zero. Here, \(a=1\), \(b=-2\lambda\), and \(c=\lambda\), so \(D=b^2-4ac=4\lambda^2-4\lambda=4\lambda(\lambda-1)\). Setting \(D=0\) gives \(\lambda=0\) or \(\lambda=1\). Hence, option A is correct. Option B is incomplete because it omits \(\lambda=0\). Exam tip: For equal roots, immediately apply the condition \(D=0\).

The coefficients are \\(a=\alpha+1\\), \\(b=-2\alpha\\), and \\(c=\alpha\\). Therefore, the discriminant is \\(D=b^2-4ac=4\alpha^2-4\alpha(\alpha+1)=-4\alpha\\). For real roots, \\(D\ge0\\), so \\(-4\alpha\ge0\\), which gives \\(\alpha\le0\\). The value \\(\alpha=0\\) is included, whereas \\(\alpha=-1\\) is already excluded because it would make the quadratic coefficient zero. Exam tip: In parameter-based quadratic questions, apply \\(D\ge0\\) and also verify that the coefficient of \\(x^2\\) is non-zero.

Here (D=4(2t-1)^2-4(t^2+2)=4(3t^2-4t-1)). From (D<0), the interval between the two boundary roots is obtained.

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=1\\), \\(b=2(m-2)\\), and \\(c=m^2-3m+4\\). Therefore, \\(D=4(m-2)^2-4(m^2-3m+4)=-4m\\). Setting \\(D=0\\) gives \\(m=0\\). Exam tip: Whenever equal roots are mentioned, immediately use \\(b^2-4ac=0\\).

Here, \(a=1\), \(b=2(m-2)\), and \(c=m^2-3m+4\). Thus, the discriminant is \(D=b^2-4ac=4(m-2)^2-4(m^2-3m+4)=-4m\). For the equation to have no real roots, \(D<0\), so \(-4m<0\), which gives \(m>0\). Note that when \(m=0\), \(D=0\), giving two equal real roots. Exam tip: For a quadratic equation with no real roots, directly apply the condition \(D<0\).

A quadratic equation has equal roots when its discriminant \(D=b^2-4ac\) is zero. Here, \(a=3\), \(b=-2(2k+1)\), and \(c=(k+1)^2\). Thus, \(D=4(2k+1)^2-12(k+1)^2=4(k^2-2k-2)\). Setting this equal to zero gives \(k^2-2k-2=0\), so \(k=1\pm\sqrt{3}\). The values in option B result from incorrect factorisation. Exam tip: For equal roots, immediately apply the condition \(D=0\).

A quadratic equation has two distinct real roots only when its discriminant satisfies \(D>0\). Here, \(a=3\), \(b=-2(2k+1)\), and \(c=(k+1)^2\). Thus, \(D=b^2-4ac=4(2k+1)^2-12(k+1)^2=4(k^2-2k-2)\). Therefore, \(k^2-2k-2>0\), or equivalently \((k-1)^2>3\), which gives \(k<1-\sqrt{3}\) or \(k>1+\sqrt{3}\). At the endpoint values, \(D=0\), so the roots are equal. Exam tip: for two distinct real roots, first impose the condition \(D>0\).

Here, \(a=1\), \(b=-2p\), and \(c=p^2-q^2\). Therefore, the discriminant is \(D=b^2-4ac=4p^2-4(p^2-q^2)=4q^2\). Since \(q\neq 0\), we have \(D=4q^2>0\), so the equation has two real and distinct roots. In fact, the roots are \(p+q\) and \(p-q\). Option B is incorrect because equal roots require \(D=0\). Exam tip: For a quadratic equation, \(D>0\) always indicates two real and distinct roots.