Nature of Roots

For the given quadratic equation, \(A=a-1\), \(B=2a\), and \(C=a+3\). Equal roots require the discriminant \(D=B^2-4AC\) to be zero. Thus, \(D=(2a)^2-4(a-1)(a+3)=4(3-2a)\). Setting \(D=0\) gives \(3-2a=0\), so \(a=\frac{3}{2}\). The value \(a=1\) is not valid because it makes \(A=0\), so the equation is no longer quadratic. Exam tip: For equal-root questions, set the discriminant to zero and then verify that the coefficient of \(x^2\) is non-zero.

For a quadratic equation to have real roots, its discriminant must satisfy D≥0. Here a=1, b=2(λ−1), and c=λ²+1. Therefore D=[2(λ−1)]²−4(1)(λ²+1)=4(λ−1)²−4(λ²+1). Expanding and simplifying gives D=4(λ²−2λ+1−λ²−1)=−8λ. The condition for real roots is −8λ≥0. Dividing by the negative number −8 reverses the inequality, giving λ≤0. Equality λ=0 is included because it produces D=0 and hence equal real roots. Thus option A is correct. Option B has the opposite inequality, option C is unnecessarily restrictive, and option D includes values such as λ=1/2 for which D<0 and the roots are not real.

Here, \(a=1\), \(b=-2(\alpha+2)\), and \(c=\alpha^2\). For real and distinct roots, the discriminant must satisfy \(D>0\). Thus, \(D=[-2(\alpha+2)]^2-4(1)(\alpha^2)=16(\alpha+1)\). Hence, \(16(\alpha+1)>0\), giving \(\alpha>-1\). When \(\alpha=-1\), \(D=0\), so the roots are real but equal, not distinct. Exam tip: For real and distinct roots, always use the condition \(D>0\).

A quadratic equation has equal roots when its discriminant is zero. Here, \(a=1\), \(b=2h-1\), and \(c=h^2\), so \(D=(2h-1)^2-4h^2=1-4h\). Setting \(D=0\) gives \(1-4h=0\), hence \(h=\frac{1}{4}\). Remember that equal roots always require a zero discriminant; for example, \(h=\frac{1}{2}\) gives \(D=-1\), so it is not correct.

A quadratic equation \(ax^2+bx+c=0\) has equal roots when its discriminant \(D=b^2-4ac\) is zero. Here, \(a=1\), \(b=n+2\), and \(c=2n\), so \(D=(n+2)^2-8n=n^2-4n+4=(n-2)^2\). Thus, \((n-2)^2=0\), giving \(n=2\). Substitution produces \(x^2+4x+4=0\), or \((x+2)^2=0\), confirming equal roots. Exam tip: For equal-root questions, set the discriminant equal to zero first.

For a quadratic equation, the discriminant is D = b² − 4ac. Here, a = 4, b = −4(k + 1), and c = k² + 2k. Thus, D = 16(k + 1)² − 16(k² + 2k) = 16. Since D = 16 > 0 for every real value of k, the equation has two distinct real roots for every real k. Therefore, option A is correct. Exam tip: In parameter-based quadratic questions, simplify the discriminant completely before deciding the condition on the parameter.

Here, \(a=9\), \(b=-6(k-1)\), and \(c=(k-1)^2\). Therefore, the discriminant is \(D=b^2-4ac=[-6(k-1)]^2-4(9)(k-1)^2=0\). Hence, for every real value of \(k\), the roots are real and equal. The equation can also be written as \([3x-(k-1)]^2=0\). Even when \(k=1\), it becomes \(9x^2=0\), which is still a quadratic equation. Exam tip: When \(D=0\), the roots are real and equal.

In the first equation, (D=100-92=8>0), and (8) is not a perfect square. So the roots are real, irrational and distinct.

A quadratic equation has equal roots when its discriminant is \(D=0\). Thus, \((m-5)^2=0\), which gives \(m-5=0\) and hence \(m=5\). If \(m=-5\), the discriminant is \((-10)^2=100\), so the roots are not equal. Exam tip: The square of a real number is zero only when the number itself is zero.

Since \(r\) is real, \(r^2\ge 0\). Therefore, \(D=2r^2+3\ge 3>0\). For a quadratic equation, \(D>0\) indicates two real and distinct roots. Equal roots require \(D=0\), so option B is incorrect. Exam tip: remember the three discriminant tests: \(D>0\) for real and distinct roots, \(D=0\) for equal roots, and \(D<0\) for non-real roots.

For every real ctc, c(t^2\ge 0)c, so c(t^2+4>0)c and therefore cD=-(t^2+4)<0c. A negative discriminant means that a quadratic equation has no real roots; its roots are complex. Hence, the correct answer is 0. Exam tip: cD<0c, cD=0c and cD>0c indicate 0, 1 and 2 real roots, respectively.

For a quadratic with real coefficients, every non-real root occurs with its complex conjugate. Hence one root cannot be real while the other is non-real. Exam tip: use the sign of the discriminant to check root type.

Here, \(a=1\), \(b=2p\), and \(c=p^2-1\). Therefore, the discriminant is \(D=b^2-4ac=(2p)^2-4(p^2-1)=4\). Since \(D>0\) for every real value of \(p\), both roots are always real and distinct. Option B is incorrect because equal roots require \(D=0\). Exam tip: For a quadratic involving a parameter, simplify the discriminant before testing particular parameter values.

For this quadratic equation, \(a=1\), \(b=-2p\), and \(c=p^2\). Therefore, the discriminant is \(\Delta=b^2-4ac=(-2p)^2-4(1)(p^2)=0\). Hence the roots are always real and equal. In fact, the equation can be written as \((x-p)^2=0\), so both roots are \(x=p\). Option B is incorrect because \(\Delta=0\) does not give distinct roots. Exam tip: \(\Delta=0\) indicates equal real roots.

Here, \(a=1\), \(b=2p\), and \(c=p^2+1\). Thus, the discriminant is \(D=b^2-4ac=(2p)^2-4(1)(p^2+1)=-4\), which remains negative for every real \(p\). Therefore, the equation has no real roots. Option D is also incorrect because putting \(p=0\) gives \(x^2+1=0\), which has no real roots. Exam tip: For a quadratic equation, \(D<0\) means the roots are non-real conjugates.

For the quadratic equation, \(A=2\), \(B=-4a\), and \(C=2a^2-3\). Its discriminant is \(D=B^2-4AC=(-4a)^2-4(2)(2a^2-3)=16a^2-16a^2+24=24\). Since \(D=24>0\) for every real value of \(a\), the roots are always real and distinct. Hence option A is correct; equal roots would require \(D=0\). Exam tip: in parameter-based questions, simplify the discriminant fully and check whether its sign depends on the parameter.

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.