Nature of Roots

Here, \(A=3\), \(B=-6a\), and \(C=3a^2+2\). Therefore, the discriminant is \(D=B^2-4AC=(-6a)^2-4(3)(3a^2+2)=36a^2-36a^2-24=-24\). Since \(D<0\), the equation has no real roots for every real value of \(a\). Option D is incorrect because the discriminant remains fixed at \(-24\), independent of \(a\). Exam tip: For a quadratic equation, \(D<0\) indicates that the roots are not real.

For a quadratic equation \(Ax^2+Bx+C=0\) to have equal roots, its discriminant must be zero: \(D=B^2-4AC=0\). Here, \(A=1\), \(B=-2(a+b)\), and \(C=(a-b)^2\). Thus, \(D=4(a+b)^2-4(a-b)^2=16ab\). Therefore, \(16ab=0\), giving \(ab=0\). Options B and C describe only particular cases, whereas option A gives the necessary and sufficient relation. Exam tip: For equal roots, immediately apply \(D=0\).

For equal roots, the discriminant must be zero. Here,
\Delta=[-2(a+b)]^2-4(1)(4ab)=4(a+b)^2-16ab=4(a-b)^2
. Thus,
\Delta=0
gives
(a-b)^2=0
, so
a=b
. Options C and D can also hold in the special case
a=b=0
, but they are not necessary relations. Exam tip: For equal roots of a quadratic equation, first set
\Delta=0
.

A quadratic equation has equal roots when its discriminant is zero. Here, \(D=(a+b)^2-4ab=a^2-2ab+b^2=(a-b)^2\). Since a and b are real, \((a-b)^2=0\) only when \(a=b\). Therefore, option A is correct. Exam tip: set \(D=0\) first; conditions such as \(a+b=0\) or \(ab=1\) do not generally imply equal roots.

Here, \(A=1\), \(B=-(a+b)\), and \(C=ab\). Therefore, the discriminant is \(\Delta=B^2-4AC=(a+b)^2-4ab=(a-b)^2\). Since \(a\ne b\), \(\Delta>0\), so the roots are real and distinct. In fact, the equation factors as \((x-a)(x-b)=0\), giving the roots \(a\) and \(b\). Option B would apply only when \(a=b\), and the roots need not be irrational. Exam tip: For a quadratic equation, \(\Delta>0\) indicates two real and distinct roots.

For a quadratic equation, the discriminant is \(D=b^2-4ac\). Here, \(a_1=1\), \(b=2(a+1)\), and \(c=a^2+2a+5\). Thus, \(D=[2(a+1)]^2-4(a^2+2a+5)=-16<0\). Therefore, for every real value of \(a\), the equation has no real roots. Exam tip: When \(D<0\), the roots are non-real and conjugate to each other.

For a quadratic equation to have real and equal roots, its discriminant must be zero. Here, \(a=1\), \(b=-2m\), and \(c=3m\), so \(D=b^2-4ac=(-2m)^2-4(1)(3m)=4m(m-3)\). Setting \(D=0\) gives \(m=0\) or \(m=3\). Both values are valid because the coefficient of \(x^2\) remains non-zero. Exam tip: For equal real roots, immediately use the condition \(D=0\).

A quadratic equation has two real and distinct roots only when its discriminant satisfies \(D>0\). Here, \(a=1, b=-2m, c=3m\), so \(D=b^2-4ac=4m^2-12m=4m(m-3)\). Therefore, \(4m(m-3)>0\), which gives \(m<0\) or \(m>3\). At \(m=0\) or \(m=3\), the discriminant is zero, so the roots are equal rather than distinct. Exam tip: For real and distinct roots, apply \(D>0\).

A quadratic equation has no real roots when its discriminant satisfies \(D<0\). Here, \(a=1\), \(b=-2m\), and \(c=3m\), so \(D=b^2-4ac=4m^2-12m=4m(m-3)\). Thus, \(4m(m-3)<0\), which gives \(0<m<3\). At \(m=0\) or \(m=3\), \(D=0\), so the equation has real and equal roots rather than no real roots. Exam tip: remember that \(D<0\), \(D=0\), and \(D>0\) correspond respectively to no real roots, equal real roots, and two distinct real 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 has no real roots exactly when its discriminant is negative. In x²+2(a−2)x+2a+5=0, the coefficients are A=1, B=2(a−2), and C=2a+5. Therefore 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 equation has roots a=3±√10. Since the coefficient of a² is positive, the expression is negative strictly between these roots. Hence 3−√10<a<3+√10, making option A correct. Equality gives D=0, and outside the interval gives D>0.

The discriminant of a quadratic equation is \(D=b^2-4ac\). When \(D<0\), \(\sqrt{D}\) is not real, so the equation has no real roots; its roots are complex conjugates. Equal real roots require \(D=0\), so option B is incorrect. In an exam, first determine the sign of the discriminant.

Here, \(a=3\), \(b=-10\), and \(c=3\). Therefore, the discriminant is \(D=b^2-4ac=(-10)^2-4(3)(3)=100-36=64\). Since \(D\) is positive and a perfect square, the equation has two real, rational, and distinct roots. In fact, the roots are \(3\) and \(\frac{1}{3}\). Exam tip: when \(D>0\) and is a perfect square, the roots are rational and distinct.

Here, \(a=5\), \(b=-6\), and \(c=2\). The discriminant is \(D=b^2-4ac=(-6)^2-4(5)(2)=36-40=-4\). Since \(D<0\), the equation has no real roots. Equal real roots occur only when \(D=0\), so option B is incorrect; the other listed discriminant values are also miscalculations. Exam tip: determine the nature of quadratic roots by checking the sign of the discriminant first.

For a quadratic equation Ax² + Bx + C = 0, the discriminant D = B² − 4AC determines the nature of its roots. Here A = 1, B = −9, and C = 20. Thus D = (−9)² − 4(1)(20) = 81 − 80 = 1. Since D is positive, the roots are real and distinct. Moreover, D = 1 is a perfect square, and with rational coefficients this means both roots are rational. Indeed, the equation factors as (x − 4)(x − 5) = 0, giving roots 4 and 5. Therefore option A is correct; D = 0 would indicate equal roots, while D < 0 would indicate no real roots.

Here, \(a=1, b=6, c=13\). The discriminant is \(D=b^2-4ac=6^2-4(1)(13)=36-52=-16\). Since \(D<0\), the equation has no real roots; its roots are complex conjugates. Option B would require \(D=0\), which is not the case here. Exam tip: For a quadratic equation, \(D<0\) means that there are no real roots.

Here, \(a=4, b=-20, c=25\). The discriminant is \(D=b^2-4ac=(-20)^2-4(4)(25)=0\), so the two roots are equal. Factoring gives \((2x-5)^2=0\), hence \(2x-5=0\) and the repeated root is \(x=\frac{5}{2}\). Exam tip: When \(D=0\), the equal root can also be found directly using \(x=\frac{-b}{2a}\).

The discriminant D=B²−4AC determines whether the roots are real, equal, or non-real. For x²−5x+5=0, A=1, B=−5, and C=5. Thus D=(−5)²−4(1)(5)=25−20=5. Since D>0, there are two real and distinct roots. Moreover, 5 is not a perfect square, so √5 is irrational. The quadratic formula gives x=[5±√5]/2, and both values are irrational. Therefore option A correctly combines reality, distinctness, and irrationality. Option B uses an incorrect discriminant value of 25; option C would require D=0, and option D would require D<0. Hence A is unambiguously correct.