Nature of Roots

Equal roots require the discriminant to be zero. Here, \(a=1\), \(b=-(u+6)\), and \(c=6u\). Thus, \(D=b^2-4ac=(u+6)^2-24u=(u-6)^2\). Setting \(D=0\) gives \(u=6\). Exam tip: For equal roots, immediately apply the condition \(D=0\).

The governing concept is that a quadratic has real roots when its discriminant is non-negative. Here a = p − 2, b = −2(p + 2), and c = p + 6. Compute D = [−2(p + 2)]² − 4(p − 2)(p + 6) = 4(p + 2)² − 4(p² + 4p − 12) = 4(p² + 4p + 4 − p² − 4p + 12) = 64. This calculation shows D is always positive, not 40 − 8p. Therefore, for every p ≠ 2, the equation is genuinely quadratic and has two real distinct roots. None of the supplied options states this correctly: A unnecessarily restricts p, B is opposite, C violates p ≠ 2, and D is the only option that says every p ≠ 2. Hence option D, not the supplied key A, is correct.

For equal roots, the discriminant must satisfy \(D=b^2-4ac=0\). Here, \(a=q+3\), \(b=-2(q-2)\), and \(c=q\). Therefore, \(D=4(q-2)^2-4q(q+3)=4(4-7q)\). Setting \(D=0\) gives \(4-7q=0\), so \(q=\frac{4}{7}\). The value \(q=-3\) in option D makes the coefficient of \(x^2\) zero, so the equation is no longer quadratic. Exam tip: For equal-root questions, set \(D=0\) and also verify that \(a\ne0\).

Here (D=4(v+2)^2-4(4v+11)) must be expanded carefully. Always verify the middle term before solving the interval.

A quadratic equation has no real roots when its discriminant satisfies \(D<0\). Here, \(D=[2(v+2)]^2-4(4v+11)=4(v^2-7)\). Thus \(4(v^2-7)<0\), or \(v^2<7\), which gives \(-\sqrt{7}<v<\sqrt{7}\). In option B, \(D>0\), so there are two distinct real roots; in option C, \(D=0\), so there is one repeated real root. Exam tip: For ‘no real roots,’ directly apply the condition \(D<0\).

The discriminant is \(D=[-2(a+b)]^2-4(1)(3ab)=4(a^2-ab+b^2)\). Since \(a^2-ab+b^2=\frac{1}{2}[(a-b)^2+a^2+b^2]\geq 0\), we have \(D\geq 0\), so the roots are real for every pair of real values of \(a\) and \(b\). Option B is not necessary; for example, when \(a=1,b=-1\), \(ab<0\) but the roots are still real. Equal roots require \(D=0\), which occurs only when \(a=b=0\), so option D is incorrect. Exam tip: For a quadratic equation, determine the nature of its roots by checking the sign of the discriminant.

A quadratic equation has real roots when its discriminant satisfies \(D\geq0\). Here, \(D=[-2(a-2b)]^2-4(a+2b)^2=4\{(a-2b)^2-(a+2b)^2\}=-32ab\). Therefore, \(-32ab\geq0\), which gives \(ab\leq0\). Option D is only a sufficient condition: \(a+2b=0\) implies \(ab\leq0\), but it is not necessary. Exam tip: For questions about the nature of quadratic roots, begin by calculating the discriminant and imposing \(D\geq0\).

The discriminant is \(D=b^2-4ac\). Since \(a<0\) and \(c>0\), we have \(ac<0\), so \(-4ac>0\) and \(D=b^2+(-4ac)>0\). Therefore, the equation has two real and distinct roots. Equal roots require \(D=0\), while no real roots require \(D<0\), neither of which is possible here. Exam tip: Whenever \(ac<0\), the roots are automatically real and distinct.

For the quadratic equation \(ax^2+bx+c=0\), here \(b=0\) and \(a<0, c<0\), so \(ac>0\). Therefore, the discriminant is \(D=b^2-4ac=0-4ac<0\), which means that the equation has no real roots. Option B would require \(D=0\), which is not the case here. Exam tip: if \(D<0\), the roots are not real.

For a quadratic equation, the discriminant is \\(D=b^2-4ac\\). Here, \\(a=8, b=-4\sqrt{10}, c=5\\), so \\(D=(-4\sqrt{10})^2-4(8)(5)=160-160=0\\). Therefore, the roots are real and equal. In fact, the repeated root is \\(x=\frac{\sqrt{10}}{4}\\), which is irrational; however, option D is still incorrect because it says the roots are distinct. Exam tip: \\(D=0\\) always indicates two equal real roots.

The governing concept is the discriminant D = b² − 4ac. For 5x² − 3√5x + 6 = 0, the coefficients are a = 5, b = −3√5, and c = 6. Hence b² = (−3√5)² = 9 × 5 = 45, while 4ac = 4 × 5 × 6 = 120. Therefore D = 45 − 120 = −75. A negative discriminant means the equation has no real roots; its roots are a complex conjugate pair. D = 0 would describe equal real roots, while positive values would indicate two distinct real roots. Consequently option A is correct. The numerical discriminants quoted in options B, C, and D are also inconsistent with the actual calculation.

A quadratic equation has two real and distinct roots only when its discriminant is positive. Here, \(a=1\), \(b=-2\theta\), and \(c=3\theta\), so \(D=b^2-4ac=4\theta^2-12\theta=4\theta(\theta-3)\). Therefore, \(4\theta(\theta-3)>0\), which gives \(\theta<0\) or \(\theta>3\). At \(\theta=0\) and \(\theta=3\), the roots are equal, while for \(0<\theta<3\), the roots are not real. Exam tip: For two real and distinct roots, always impose \(D>0\).

For equal roots, the discriminant \(D=b^2-4ac\) must be zero. Here, \(a=1\), \(b=-2\theta\), and \(c=3\theta\), so \(D=(-2\theta)^2-4(1)(3\theta)=4\theta(\theta-3)\). Thus, \(4\theta(\theta-3)=0\) gives \(\theta=0\) or \(\theta=3\). Option B is incomplete because it omits \(\theta=0\). Exam tip: For equal roots, set the discriminant directly to zero and check every resulting parameter value.

Here (D=4\alpha^2-4(\alpha+3)(\alpha-2)=24-4\alpha). For real roots (\alpha\leq3), and for a quadratic (\alpha\neq-3).

For a quadratic equation, real roots do not exist when the discriminant \(D<0\). Here, \(D=[-2(4t+1)]^2-4(7t^2+2t+5)=4(9t^2+6t-4)\). Thus, \(9t^2+6t-4<0\). Its zeros are \(\frac{-1-\sqrt5}{3}\) and \(\frac{-1+\sqrt5}{3}\), and since the quadratic has a positive leading coefficient, the expression is negative between these zeros. Hence option A is correct. At the endpoints, \(D=0\), giving equal real roots, so the endpoints must be excluded. Exam tip: for no real roots, always use the strict condition \(D<0\), not \(D\leq0\).

For a quadratic equation \(ax^2+bx+c=0\) to have equal roots, its discriminant must be zero: \(D=b^2-4ac=0\). Here, \(a=1\), \(b=2(m-5)\), and \(c=m^2-9m+24\). Thus, \(D=4(m-5)^2-4(m^2-9m+24)=4(1-m)\). Setting \(D=0\) gives \(4(1-m)=0\), so \(m=1\). Exam tip: For equal-root questions, begin with \(D=0\) and simplify the discriminant carefully.

Here, a = 1, b = 2(m − 5), and c = m² − 9m + 24. Therefore, the discriminant is D = b² − 4ac = 4(m − 5)² − 4(m² − 9m + 24) = 4(1 − m). For the equation to have no real roots, D must be less than zero. Thus, 4(1 − m) < 0 gives m > 1. When m = 1, D = 0, so the equation has two equal real roots; hence option C is incorrect. Exam tip: For a quadratic equation, remember that no real roots occur when D < 0.

Here \(a=1, b=2p\), and \(c=p^2+1\). Thus \(\Delta=b^2-4ac=4p^2-4(p^2+1)=-4<0\), so there are no real roots. In exams, check the sign of the discriminant first.

The governing concept is the discriminant of a quadratic equation ax² + bx + c = 0. Its value is D = b² − 4ac. Here a = 1, b = −2p and c = p² − 64. Therefore, D = (−2p)² − 4(1)(p² − 64) = 4p² − 4p² + 256 = 256. This is positive for every real value of p, so the roots are real and distinct. Moreover, 256 = 16² is a perfect square, and the coefficients are real, so the roots are rational. In fact, the roots are p + 8 and p − 8. Thus option A is correct. Option B would require D = 0, option C would require D < 0, and option D is wrong because the discriminant is a perfect square.

Here, \(a=1\), \(b=-2p\), and \(c=p^2+36\). Therefore, the discriminant is \(D=b^2-4ac=(-2p)^2-4(p^2+36)=-144\), which remains negative for every real value of \(p\). Hence, the equation has no real roots. Equal real roots would require \(D=0\), whereas here \(D<0\). Exam tip: For a quadratic equation, \(D<0\) immediately indicates that there are no real roots.