Nature of Roots

For a quadratic polynomial ax² + bx + c, the discriminant D = b² − 4ac determines the nature of its zeroes. Here a = 3, b = −12 and c = 6. Therefore, D = (−12)² − 4(3)(6) = 144 − 72 = 72. Because D is positive, the polynomial has two distinct real zeroes. However, 72 is not a perfect square, so √D is irrational. Using the quadratic formula, the zeroes are [12 ± √72]/6 = 2 ± √2, which confirms that both are irrational and distinct. Thus option C is correct. Option A would require D = 0, option B would require a positive perfect-square discriminant, and option D would require D < 0.

The correct answer is A. For a quadratic equation written as ax² + bx + c = 0, the discriminant is D = b² − 4ac. The first step is to identify a, b and c carefully, including the sign of b. In option A, a = 1, b = 1 and c = 4, so D = 1² − 4(1)(4) = 1 − 16 = −15. In option B, a = 1, b = 4 and c = 1, giving D = 16 − 4 = 12, not −15. In option C, a = 2, b = 1 and c = 1, giving D = 1 − 8 = −7. In option D, a = 1, b = −5 and c = 8, giving D = (−5)² − 32 = −7. Thus only A satisfies the condition. A negative discriminant means the equation has no real roots. Memory cue: first write a, b and c separately, then use b² − 4ac; never forget a when it is not 1.

Let the two roots be 3 and r. Since the equation is x² + ax + 12 = 0, its leading coefficient is 1 and Vieta’s product relation gives 3r = 12. Hence r = 4. The sum of the roots is therefore 3 + 4 = 7. Vieta’s sum relation says that the sum equals −a, because the coefficient of x² is 1. Thus −a = 7 and a = −7, so option A is correct. Direct substitution confirms the value: putting x = 3 into the equation gives 9 + 3a + 12 = 0, or 3a + 21 = 0, which again yields a = −7. The other choices either use a negative second root, the wrong sign for a, or values that fail the product and substitution conditions.

The governing concept is the discriminant criterion for the nature of roots. For a quadratic ax^2+bx+c=0, the roots are non-real when the discriminant D=b^2−4ac is strictly less than zero. Here a=5, b=−4 and c=λ, so D=(−4)^2−4(5)(λ)=16−20λ. Requiring D<0 gives 16−20λ<0, hence −20λ<−16. Dividing by the negative number −20 reverses the inequality, yielding λ>16/20=4/5. Therefore option C is correct. At λ=4/5, D=0 and the equation has one repeated real root, so equality is excluded. If λ<4/5, then D>0 and there are two distinct real roots; this rules out A, B and D.

The nature of the roots is determined by the discriminant \(\Delta=b^2-4ac\). Real and equal roots occur only when \(\Delta=0\), which gives \(b^2=4ac\). If \(b^2>4ac\), the roots are real and distinct; if \(b^2<4ac\), they are non-real. Exam tip: for equal roots, immediately set the discriminant equal to zero.

When \(D>0\), the roots are real and distinct. With integer coefficients, if \(D\) is not a perfect square, \(\sqrt D\) is irrational, so both roots are irrational. Exam tip: check the sign of \(D\) first, then test whether it is a perfect square.

For the quadratic equation, \\(a=1\\), \\(b=-2(k+1)\\), and \\(c=k^2\\). Therefore, the discriminant is \\(D=b^2-4ac=4(k+1)^2-4k^2=4(2k+1)\\). Distinct real roots require \\(D>0\\), so \\(2k+1>0\\), which gives \\(k>-\\frac{1}{2}\\). At \\(k=-\\frac{1}{2}\\), \\(D=0\\), so the roots are equal rather than distinct. Exam tip: determine the sign of the discriminant before classifying the roots.

Here, \(a=1\), \(b=2(m-1)\), and \(c=m+5\). For no real roots, the discriminant must satisfy \(D<0\). Thus, \(D=4(m-1)^2-4(m+5)=4(m^2-3m-4)=4(m-4)(m+1)\). Therefore, \((m-4)(m+1)<0\), giving \(-1<m<4\). In option B, the discriminant is positive, so there are two real roots; at the endpoint values in option C, the roots are real and equal. Exam tip: For a quadratic equation, determine the nature of the roots by checking the sign of its discriminant.

A quadratic equation has equal roots when its discriminant satisfies \\(D=b^2-4ac=0\\). Here, \\(a=1, b=-2(k-2), c=k^2-4\\), so \\(D=4(k-2)^2-4(k^2-4)=16(2-k)\\). Setting \\(D=0\\) gives \\(16(2-k)=0\\), hence \\(k=2\\). As a check, when \\(k=2\\), the equation becomes \\(x^2=0\\), which has the repeated root \\(x=0\\). Exam tip: For equal roots, directly apply \\(b^2-4ac=0\\).

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(\Delta=b^2-4ac\). Here, \(a=3, b=2\sqrt{3}, c=1\), so \(\Delta=(2\sqrt{3})^2-4(3)(1)=12-12=0\). A zero discriminant means that the two roots are real and equal. The repeated root is \(x=-\frac{\sqrt{3}}{3}\); therefore, the roots are not distinct. Exam tip: whenever \(\Delta=0\), select ‘real and equal roots’.

Here (D=(2\sqrt{5})^2-4(1)(1)=16>0). The roots are (\sqrt{5}\pm2), so they are irrational and distinct.

For a quadratic equation \(ax^2+bx+c=0\) to have real and equal roots, its discriminant must be zero: \(D=b^2-4ac=0\). Here, \(a=2\), \(b=p\), and \(c=8\). Thus, \(p^2-4(2)(8)=0\), so \(p^2=64\) and \(p=\pm 8\). Therefore, option A is correct. Exam tip: when taking the square root of \(p^2=64\), include both the positive and negative values; hence option C is incomplete.

Here, \\(a=p-2\\), \\(b=4\\), and \\(c=1\\). An equation has equal roots when its discriminant \\(D=b^2-4ac\\) is zero. Thus, \\(D=16-4(p-2)=24-4p=0\\), giving \\(p=6\\). The value \\(p=2\\) is not valid because it eliminates the quadratic term, so the equation is no longer quadratic. Exam tip: For equal roots, set \\(D=0\\) and also verify that \\(a\ne0\\).

Here, \(a=k\), \(b=-6\), and \(c=k\). Therefore, the discriminant is \(D=b^2-4ac=36-4k^2\). Real and distinct roots require \(D>0\), giving \(36-4k^2>0\), or \(k^2<9\). In addition, the equation must remain quadratic, so its leading coefficient must satisfy \(k\ne0\). Hence the correct condition is \(k^2<9\) and \(k\ne0\). Option B is incomplete because it allows \(k=0\), which makes the equation linear. Exam tip: for distinct real roots, impose \(D>0\) and also verify that the leading coefficient is non-zero.

A quadratic equation \(Ax^2+Bx+C=0\) has equal roots only when its discriminant \(D=B^2-4AC\) is zero. Here, \(A=1\), \(B=-(a+3)\), and \(C=3a\), so \(D=(a+3)^2-12a=(a-3)^2\). Thus, \((a-3)^2=0\) gives \(a=3\), making option A correct. For comparison, when \(a=0\), \(D=9\), so the roots are not equal. Exam tip: For equal roots, set the discriminant equal to zero.

For a quadratic equation \(ax^2+bx+c=0\), equal roots occur when the discriminant \(D=b^2-4ac\) is zero. Here, \(a=1\), \(b=-(r+4)\), and \(c=4r\), so \(D=(r+4)^2-16r=(r-4)^2\). Setting \(D=0\) gives \((r-4)^2=0\), hence \(r=4\). Exam tip: whenever a question asks for equal roots, begin with the condition \(D=0\).

A quadratic equation has real roots when its discriminant satisfies \(D\ge0\). Here, \(D=[2(a+1)]^2-4(a^2+3)=8(a-1)\). Therefore, \(8(a-1)\ge0\), giving \(a\ge1\). Option C, \(a>3\), is sufficient but unnecessarily restrictive; it is not the complete condition. Exam tip: For a parameter-based quadratic equation, begin by setting its discriminant greater than or equal to zero.

A quadratic equation has real and equal roots only when its discriminant is zero. Here, \(D=[-2(a-3)]^2-4(a^2-9)=24(3-a)\). Therefore, \(24(3-a)=0\), giving \(a=3\). As a check, substituting \(a=3\) reduces the equation to \(x^2=0\), whose roots are equal. Exam tip: For equal roots, set the discriminant \(D\) equal to zero.

A quadratic equation has real and distinct roots only when its discriminant satisfies \(D>0\). Here, \(a=1\), \(b=-2s\), and \(c=s+2\), so \(D=(-2s)^2-4(1)(s+2)=4(s-2)(s+1)\). Thus, \((s-2)(s+1)>0\), which gives \(s<-1\) or \(s>2\). For \(-1<s<2\), \(D<0\), while \(s=-1\) or \(s=2\) gives \(D=0\). Exam tip: For real and distinct roots, always check that \(D>0\).

A quadratic equation has no real roots when its discriminant satisfies \(D<0\). Here, \(a=1, b=2t, c=t+4\), so \(D=(2t)^2-4(1)(t+4)=4(t^2-t-4)\). Thus, \(t^2-t-4<0\), whose zeros are \(\frac{1-\sqrt{17}}{2}\) and \(\frac{1+\sqrt{17}}{2}\). Since the corresponding parabola opens upward, the inequality holds between these two zeros. Exam tip: For \(D<0\), choose the interval between the discriminant’s zeros and exclude the endpoints.

Here (D=(3q+1)^2-8q=9q^2-2q+1). Its own discriminant ((-2)^2-4(9)(1)<0), so it is always positive.

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.