Nature of Roots

Here \\(A=1\\), \\(B=-(2a+1)\\), and \\(C=a(a+1)\\). Therefore, the discriminant is \\(D=B^2-4AC=(2a+1)^2-4a(a+1)=1\\). Since \\(D>0\\), the roots are real and distinct. In fact, the roots are \\(a\\) and \\(a+1\\). They cannot always be called rational or irrational, because that depends on the value of \\(a\\). Exam tip: simplify the discriminant first and then determine its sign.

The equation can be factorised as \(x^2-(a+b)x+ab=(x-a)(x-b)=0\). Hence, its roots are \(x=a\) and \(x=b\). Since a and b are real and unequal, the roots are distinct and real. Equal roots would occur only when \(a=b\), and the roots need not be irrational. Exam tip: Use the discriminant \(D=(a-b)^2\); when \(a\neq b\), \(D>0\), which indicates two distinct real roots.

A quadratic equation has no real roots when its discriminant satisfies \(D<0\). Here, \(D=[-2(a-1)]^2-4(a^2+1)=4(a-1)^2-4(a^2+1)=-8a\). Therefore, \(-8a<0\), which gives \(a>0\). For \(a=0\), \(D=0\), so the equation has two real and equal roots. Exam tip: whenever ‘no real roots’ is stated, immediately apply the condition \(D<0\).

When (D) is negative, no real root exists. Graphically, the parabola does not cut the (x)-axis.

Here, \(a=2\), \(b=-6\sqrt{2}\), and \(c=9\). Therefore, the discriminant is \(\Delta=b^2-4ac=(-6\sqrt{2})^2-4(2)(9)=72-72=0\). When \(\Delta=0\), the quadratic equation has two real and equal roots. In fact, the repeated root is \(x=\frac{3\sqrt{2}}{2}\). Hence, option A is correct; options B and D require a positive discriminant, whereas the discriminant here is zero. Exam tip: To determine the nature of roots, first check the sign of \(\Delta\).

Here, \(a=3\), \(b=-4\sqrt{3}\), and \(c=5\). Therefore, the discriminant is \(\Delta=b^2-4ac=(-4\sqrt{3})^2-4(3)(5)=48-60=-12\). Since \(\Delta<0\), the equation has no real roots. Option B is incorrect because equal real roots require \(\Delta=0\). Exam tip: For a quadratic equation, a negative discriminant always indicates that no real roots exist.

Here, \\(a=5\\), \\(b=-2\\sqrt{30}\\), and \\(c=6\\). Therefore, the discriminant is \\(\\Delta=b^2-4ac=(-2\\sqrt{30})^2-4(5)(6)=120-120=0\\). Hence, the roots are real and equal. Option D is incorrect because \\(\\Delta>0\\) would indicate distinct roots. Exam tip: For a quadratic equation, \\(\\Delta=0\\) always means two equal real roots.

For a quadratic equation to have real roots, its discriminant must satisfy \(D\geq 0\). Here, \(D=[-2(k+2)]^2-4(k^2+3k+7)=4(k-3)\). Therefore, \(4(k-3)\geq 0\), which gives \(k\geq 3\). Option B is incorrect because \(k=3\) also gives real, equal roots. Exam tip: For questions about real roots, begin by applying \(D\geq 0\).

The governing concept is the discriminant criterion for a quadratic equation. For ax² + bx + c = 0, the roots are non-real exactly when D = b² − 4ac < 0. Here a = 1, b = −2(k − 3), and c = k² − 8k + 20. Therefore D = [−2(k − 3)]² − 4(k² − 8k + 20) = 4(k − 3)² − 4(k² − 8k + 20) = 4(2k − 11). For no real roots, 4(2k − 11) < 0. Since 4 is positive, this reduces to 2k − 11 < 0, giving k < 11/2. At k = 11/2, D = 0 and the roots are equal and real; for larger k, D > 0 and two distinct real roots occur. Thus option A is correct.

Here, \(a=1\), \(b=-(r+5)\), and \(c=5r\). Therefore, the discriminant is \(D=b^2-4ac=(r+5)^2-20r=(r-5)^2\). Two real and distinct roots require \(D>0\). Hence, \((r-5)^2>0\), which gives \(r\ne5\). In option B, \(r=5\) makes \(D=0\), so the roots are equal. Exam tip: For two distinct real roots of a quadratic equation, always check that \(D>0\).

Here, \(a=1\), \(b=-(s+4)\), and \(c=4s\). Equal roots require the discriminant \(D=b^2-4ac\) to be zero. Thus, \(D=(s+4)^2-16s=s^2-8s+16=(s-4)^2\). Therefore, \((s-4)^2=0\), giving \(s=4\). The distractor \(s=0\) is incorrect because it gives \(D=16\), not zero. Exam tip: for equal roots, immediately apply the condition \(D=0\).

Because p ≠ 1, the coefficient p − 1 of x² is nonzero, so the equation is genuinely quadratic. Real roots require a non-negative discriminant. Here a = p − 1, b = −2(p + 1), and c = p + 3. Thus D = [−2(p + 1)]² − 4(p − 1)(p + 3). Expanding, (p + 1)² = p² + 2p + 1 and (p − 1)(p + 3) = p² + 2p − 3. Hence D = 4[(p² + 2p + 1) − (p² + 2p − 3)] = 16. This is positive for every real p. Therefore two distinct real roots exist for every allowed p, namely every p except 1. Option D is correct. Option A imposes an unsupported upper bound, and p = 1 is explicitly excluded.

Equal roots occur when the discriminant of a genuine quadratic is zero. Here a = q + 2, b = −2(q − 1), and c = q. Therefore D = [−2(q − 1)]² − 4(q + 2)q = 4(q − 1)² − 4q(q + 2). Expanding the terms gives D = 4(q² − 2q + 1 − q² − 2q) = 4(1 − 4q). Set D equal to zero: 4(1 − 4q) = 0, so 1 − 4q = 0 and q = 1/4. This value satisfies the given restriction q ≠ −2, so the coefficient of x² remains nonzero. The other options leave a nonzero discriminant and therefore do not produce equal roots. Hence option A is correct.

For a quadratic equation \\(ax^2+bx+c=0\\), equal roots occur when the discriminant \\(D=b^2-4ac\\) is zero. Here, \\(a=q+2\\), \\(b=-2(q-1)\\), and \\(c=q\\). Thus, \\(D=4(q-1)^2-4q(q+2)=4(1-4q)\\). Setting \\(D=0\\) gives \\(1-4q=0\\), so \\(q=\\frac14\\). At \\(q=1\\), the discriminant is not zero, while \\(q=-2\\) removes the quadratic term and is also excluded by the condition. Exam tip: For equal roots, begin by setting the discriminant equal to zero.

For a quadratic to have no real roots, its discriminant must be negative. In this equation a = 1, b = 2(t + 1), and c = 3t + 7. Hence D = [2(t + 1)]² − 4(3t + 7) = 4(t + 1)² − 4(3t + 7) = 4(t² − t − 6) = 4(t − 3)(t + 2). The constant factor 4 is positive, so D is negative precisely when (t − 3)(t + 2) is negative. A product of two linear factors is negative between its distinct zeros, −2 and 3. Therefore −2 < t < 3, making option A correct. At either endpoint D = 0 and the roots are equal and real; outside the interval D > 0 and two distinct real roots result.

Here, \(a=1\), \(b=2(t+1)\), and \(c=3t+7\). Therefore, the discriminant is \(D=b^2-4ac=4(t+1)^2-4(3t+7)=4(t-3)(t+2)\). An equation has no real roots when \(D<0\), so \((t-3)(t+2)<0\), giving \(-2<t<3\). Option B is incorrect because it does not come from the zeros \(-2\) and \(3\); at the endpoints \(t=-2,3\), \(D=0\) and the equation has equal real roots. Exam tip: For a quadratic equation, use \(D<0\) to identify the condition for no real roots.

For the equation \\(x^2-2(a+b)x+2ab=0\\), the coefficients are 1, \\(-2(a+b)\\) and 2ab. Its discriminant is \\(D=[-2(a+b)]^2-8ab=4(a+b)^2-8ab=4(a^2+b^2)\\). Since squares are non-negative, \\(a^2+b^2\geq0\\), so D is always non-negative.

Therefore the equation always has real roots, including the possibility of equal roots when both a and b are zero. Option A expresses this correctly. The condition ab>0 is not necessary; a and b may have opposite signs or one may be zero. Also, equal roots require D=0, namely a=b=0, not merely a+b=0.

The discriminant of the quadratic equation is \(D=[-2(a-b)]^2-4(a+b)^2\). Therefore, \(D=4[(a-b)^2-(a+b)^2]=-16ab\). For real roots, \(D\geq0\), so \(-16ab\geq0\), which gives \(ab\leq0\). Option B reverses the inequality. The condition \(a+b=0\) may satisfy the requirement, but it is not the complete necessary condition. Exam tip: begin with \(D\geq0\) whenever the nature of roots is asked.

Since \(a>0\) and \(c<0\), we have \(ac<0\). Therefore, \(-4ac>0\), and for any real \(b\), the discriminant \(D=b^2-4ac=b^2+(-4ac)>0\). Hence, the equation has two real and distinct roots. Equal roots would require \(D=0\), which is impossible here. Exam tip: When \(ac<0\), the roots are always real and distinct.

Here, a=7, b=-2\(\sqrt{21}\), and c=3. Therefore, the discriminant is \(D=b^2-4ac=(-2\sqrt{21})^2-4(7)(3)=84-84=0\). Hence, the roots are real and equal. In fact, the repeated root is \(x=\frac{\sqrt{21}}{7}\). Option D incorrectly treats 84, which is only the value of \(b^2\), as the discriminant. Exam tip: \(D=0\) always indicates two equal real roots.