Nature of Roots

Here, a = 1, b = 2, and c = p² + 2. Therefore, the discriminant is D = b² − 4ac = 4 − 4(p² + 2) = −4(p² + 1). For every real p, p² + 1 > 0, so D < 0; hence the equation has no real roots. Therefore, option A is correct. Option C is incorrect because even when p = 0, D = −4, not zero. Exam tip: A quadratic equation has no real roots whenever its discriminant is negative.

For a quadratic equation to have real and distinct roots, its discriminant must satisfy \(D>0\). Here, \(a=1\), \(b=-2p\), and \(c=p^2-5p\), so \(D=b^2-4ac=4p^2-4(p^2-5p)=20p\). Thus, \(20p>0\), which gives \(p>0\). When \(p=0\), the discriminant is zero and the roots are equal, so \(p\ge0\) is not fully correct. Exam tip: For real and distinct roots, always use the condition \(D>0\).

A quadratic equation has real roots when its discriminant satisfies \(D\ge 0\). Here, \(a=1\), \(b=-2u\), and \(c=u^2+7u\), so \(D=(-2u)^2-4(1)(u^2+7u)=-28u\). Thus, \(-28u\ge0\), giving \(u\le0\). Remember that \(D=0\), which gives equal real roots, is also included.

For an equation
ax^2+bx+c=0
), equal roots require the discriminant
D=b^2-4ac
to be zero. Here,
a=m, b=m+3, c=3
, so
D=(m+3)^2-12m=(m-3)^2
. Thus,
D=0
gives
m=3
. For this value, the coefficient of
x^2
is
a=m=3
, which is non-zero, so the equation remains quadratic. Although
m=0
also makes the discriminant zero, it removes the quadratic term and therefore does not satisfy the condition. Exam tip: After setting
D=0
, always verify that the coefficient of
x^2
is not zero.

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.