Nature of Roots

For a quadratic equation ax² + bx + c = 0, the discriminant is D = b² − 4ac. In option (A), D = 2² − 4(1)(5) = −16. Since D < 0, the equation has no real roots. In option (D), D = 0, so it has one repeated real root, while options (B) and (C) have D > 0. Exam tip: Check the sign of the discriminant first to determine the nature of the roots.

For a quadratic equation 8ax^2+bx+c=09 to have equal roots, its discriminant 8D=b^2-4ac9 must be zero. Here, 8a=1,b=p,c=169, so 8D=p^2-4\times1\times16=p^2-64=09. Hence 8p^2=649 and 8p=\pm89, making option A correct. For the closest distractor, 8p=\pm49 gives a non-zero discriminant. Exam tip: whenever roots are equal, set the discriminant equal to zero first.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=1, b=-4, c=1\), so \(D=(-4)^2-4(1)(1)=12\). Since \(D>0\) and 12 is not a perfect square, the roots are real, irrational, and distinct. Option B is incorrect because \(D=16\) is not the discriminant of this equation. Exam tip: First use the sign of \(D\) to determine whether the roots are real and distinct, then check whether \(D\) is a perfect square to determine rationality.

For a quadratic equation \(ax^2+bx+c=0\), the roots are equal only when the discriminant \(D=b^2-4ac\) is zero. Therefore, the equation with \(D_1=0\) has equal roots. When \(D_2=7>0\), the roots are distinct and real, while \(D_3=-2<0\) gives no real roots. Exam tip: For equal roots, check \(D=0\) directly.

Here, \(a=1\), \(b=-8\), and \(c=16\). Thus, the discriminant is \(D=b^2-4ac=(-8)^2-4(1)(16)=64-64=0\). When \(D=0\), a quadratic equation has two equal real roots. In fact, \(t^2-8t+16=(t-4)^2\), so both roots are \(t=4\). Option B applies only when \(D>0\), which gives distinct real roots. Exam tip: To determine the nature of roots, first check the sign of the discriminant.

For equal roots, the discriminant \\(D=b^2-4ac\\) must be zero. Here, \\(a=q\\), \\(b=5\\), and \\(c=q\\), so \\(D=25-4q^2=0\\). Thus, \\(q^2=\\frac{25}{4}\\), giving \\(q=\\pm\\frac{5}{2}\\). The values \\(5\\) and \\(-5\\) in option B do not make the discriminant zero. Also, \\(q=0\\) is excluded because it would make the equation non-quadratic. Exam tip: for equal roots, set the discriminant directly equal to zero.

A quadratic equation has no real roots when its discriminant satisfies \(D<0\). Here, \(a=1\), \(b=m-2\), and \(c=1\), so \(D=(m-2)^2-4\). Thus, \((m-2)^2<4\), which gives \(-2<m-2<2\), and hence \(0<m<4\). Therefore, option A is correct. In option C, \(D=0\), which gives two equal real roots rather than no real roots. Exam tip: For ‘no real roots’, first apply the condition \(D<0\).

For equal roots, the discriminant \(D=b^2-4ac\) must be zero. Here, \(a=1\), \(b=-2(k-1)\), and \(c=k+1\). Thus, \(D=4(k-1)^2-4(k+1)=4(k^2-3k)=4k(k-3)\). Setting \(D=0\) gives \(k=0\) or \(k=3\), so option A is correct. Exam tip: whenever a quadratic equation has equal roots, begin by applying \(D=0\).

A quadratic equation has no real roots when its discriminant satisfies \(D<0\). Here, \(a=2\), \(b=k-4\), and \(c=8\), so \(D=(k-4)^2-4(2)(8)=(k-4)^2-64\). Therefore, \((k-4)^2<64\), which gives \(-8<k-4<8\), and hence \(-4<k<12\). Thus, option A is correct. In option B, \(D>0\), giving two distinct real roots, while in option C, \(D=0\), giving equal real roots. Exam tip: For the nature of roots, first calculate the discriminant using \(D=b^2-4ac\).

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(D=5^2-4(1)(7)=25-28=-3\), which is negative. Therefore, the equation has no real roots. Option A is incorrect because a negative discriminant cannot produce real roots. Exam tip: when \(D<0\), the quadratic has no real roots.

The discriminant of a quadratic equation is \(D=b^2-4ac\). When \(D<0\), \(\sqrt{D}\) is not real, so the equation has no real roots; its roots are complex conjugates. Equal real roots require \(D=0\), so option B is incorrect. In an exam, first determine the sign of the discriminant.

Here, \(a=3\), \(b=-10\), and \(c=3\). Therefore, the discriminant is \(D=b^2-4ac=(-10)^2-4(3)(3)=100-36=64\). Since \(D\) is positive and a perfect square, the equation has two real, rational, and distinct roots. In fact, the roots are \(3\) and \(\frac{1}{3}\). Exam tip: when \(D>0\) and is a perfect square, the roots are rational and distinct.

Here, \(a=5\), \(b=-6\), and \(c=2\). The discriminant is \(D=b^2-4ac=(-6)^2-4(5)(2)=36-40=-4\). Since \(D<0\), the equation has no real roots. Equal real roots occur only when \(D=0\), so option B is incorrect; the other listed discriminant values are also miscalculations. Exam tip: determine the nature of quadratic roots by checking the sign of the discriminant first.

Here, \(a=1, b=6, c=13\). The discriminant is \(D=b^2-4ac=6^2-4(1)(13)=36-52=-16\). Since \(D<0\), the equation has no real roots; its roots are complex conjugates. Option B would require \(D=0\), which is not the case here. Exam tip: For a quadratic equation, \(D<0\) means that there are no real roots.

Here, \(a=4, b=-20, c=25\). The discriminant is \(D=b^2-4ac=(-20)^2-4(4)(25)=0\), so the two roots are equal. Factoring gives \((2x-5)^2=0\), hence \(2x-5=0\) and the repeated root is \(x=\frac{5}{2}\). Exam tip: When \(D=0\), the equal root can also be found directly using \(x=\frac{-b}{2a}\).

The discriminant D=B²−4AC determines whether the roots are real, equal, or non-real. For x²−5x+5=0, A=1, B=−5, and C=5. Thus D=(−5)²−4(1)(5)=25−20=5. Since D>0, there are two real and distinct roots. Moreover, 5 is not a perfect square, so √5 is irrational. The quadratic formula gives x=[5±√5]/2, and both values are irrational. Therefore option A correctly combines reality, distinctness, and irrationality. Option B uses an incorrect discriminant value of 25; option C would require D=0, and option D would require D<0. Hence A is unambiguously correct.

Here, \(a=1\), \(b=-5\), and \(c=3\). Therefore, the discriminant is \(D=b^2-4ac=(-5)^2-4(1)(3)=25-12=13\). Since \(D>0\), the roots are real and distinct. Because 13 is not a perfect square, the roots are irrational. Hence, option C is correct. Exam tip: \(D>0\) establishes that the roots are real and distinct; check whether \(D\) is a perfect square to decide whether they are rational or irrational.

Here, \(a=3\), \(b=5\), and \(c=-2\). Therefore, the discriminant is \(D=b^2-4ac=5^2-4(3)(-2)=49\). Since \(D>0\) and 49 is a perfect square, the roots are real, rational, and distinct. In fact, the roots are \(\frac{1}{3}\) and \(-2\). Option B applies when \(D=0\), while option C applies when \(D<0\). Exam tip: If \(D>0\) and is a perfect square, the roots are real, rational, and distinct.

Use the discriminant D = B² − 4AC to classify the roots. In x² − 7x + 11 = 0, A = 1, B = −7, and C = 11. Hence D = (−7)² − 4(1)(11) = 49 − 44 = 5. Since D > 0, there are two real and distinct roots. Since 5 is not a perfect square, √5 is irrational; the quadratic formula gives roots (7 + √5)/2 and (7 − √5)/2, both irrational. Therefore option A is correct. Option B mistakes the discriminant for B² and also incorrectly calls the roots rational. Equal roots require D = 0, while no real roots require D < 0, so options C and D do not apply.

For equal roots, the discriminant \(\Delta=b^2-4ac\) must be zero. Here, \(a=1, b=k, c=25\), so \(\Delta=k^2-4(1)(25)=k^2-100\). Thus, \(k^2-100=0\), giving \(k=\pm10\), or \(k=10\) and \(k=-10\). For the closest distractor, \(k=\pm5\) gives a discriminant of \(-75\), so the roots are not equal real roots. Exam tip: For equal roots, immediately use \(b^2-4ac=0\).

For a quadratic equation \(ax^2+bx+c=0\) to have two real and distinct roots, its discriminant must satisfy \(D=b^2-4ac>0\). Here, \(a=1\), \(b=-12\), and \(c=k\), so \(D=(-12)^2-4(1)(k)=144-4k=4(36-k)\). Therefore, \(D>0\) gives \(k<36\), making option A correct. At \(k=36\), the roots are real but equal, not distinct. Exam tip: use \(D>0\) specifically for two distinct real roots.

Here, \(a=1\), \(b=4\), and \(c=k\). Therefore, the discriminant is \(D=b^2-4ac=16-4k\). A quadratic equation has no real roots when \(D<0\), so \(16-4k<0\), which gives \(k>4\). The closest distractor, \(k=4\), makes \(D=0\) and gives one repeated real root. Exam tip: Check the sign of the discriminant first to determine the nature of the roots.

For a quadratic equation \(ax^2+bx+c=0\) to have real roots, its discriminant \(D=b^2-4ac\) must satisfy \(D\geq0\). Here, \(a=2\), \(b=-3\), and \(c=p\), so \(D=(-3)^2-4(2)(p)=9-8p\). Therefore, \(9-8p\geq0\), which gives \(p\leq\frac{9}{8}\). Option C is only one particular allowed value, while option D is not necessary. Exam tip: For real roots, first apply the condition \(D\geq0\).

For equal roots, the discriminant \(D=b^2-4ac\) must be zero. Here, \(a=3\), \(b=m\), and \(c=12\), so \(m^2-4(3)(12)=0\). Therefore, \(m^2=144\), giving \(m=\pm12\), that is, \(m=12\) or \(m=-12\). Option B would give \(m^2=36\), which does not satisfy the required condition. Exam tip: For equal-root questions, start by setting the discriminant equal to zero.

For equal roots, the discriminant \(D=b^2-4ac\) must be zero. Here, \(a=k\), \(b=8\), and \(c=4\). Thus, \(D=8^2-4(k)(4)=0\), giving \(64-16k=0\) and hence \(k=4\). Values such as 8 or 16 do not make the discriminant zero. Exam tip: For a quadratic equation with equal roots, set the discriminant equal to zero.