Nature of Roots

Here, \\(a=16, b=24, c=9\\). The discriminant is \\(D=b^2-4ac=24^2-4(16)(9)=0\\), so the roots are real and equal. Also, \\(16x^2+24x+9=(4x+3)^2\\), which gives \\(4x+3=0\\) and hence the equal root \\(x=-\frac{3}{4}\\). Therefore, option B has the wrong sign, while option C gives an incorrect discriminant. Exam tip: If a quadratic is a perfect square or its discriminant is zero, its roots are equal.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=5, b=1, c=1\), so \(D=1^2-4(5)(1)=-19\). Since \(D<0\), the equation has no real roots; therefore, option A is correct. Exam tip: \(D<0\) means no real roots, \(D=0\) means equal real roots, and \(D>0\) means two distinct real roots.

For a quadratic equation, the discriminant is \\(D=b^2-4ac\\). Here, \\(a=6, b=-13, c=6\\), so \\(D=(-13)^2-4(6)(6)=169-144=25\\). Since \\(D>0\\) and 25 is a perfect square, the roots are real, rational, and distinct. In fact, the roots are \\(2/3\\) and \\(3/2\\). Option B would require \\(D=0\\). Exam tip: a positive perfect-square discriminant indicates distinct rational roots.

A quadratic equation (ax^2+bx+c=0) has equal roots when its discriminant (D=b^2-4ac) is zero. Here, (a=1,b=-14,c=k), so (-14)^2-4(1)(k)=0, giving (196-4k=0) and hence (k=49). The value 196 results from forgetting to divide by 4. Exam tip: For equal roots, immediately use (D=0).

For equal roots, the discriminant \(D=b^2-4ac\) must be zero. Here \(a=1\), \(b=k\), and \(c=36\), so \(D=k^2-4(1)(36)=k^2-144\). Therefore, \(k^2-144=0\), giving \(k^2=144\) and \(k=\pm12\). Exam tip: For equal roots of a quadratic equation, set the discriminant directly equal to zero.

Here, \(a=1\), \(b=-16\), and \(c=k\). Thus, the discriminant is \(D=b^2-4ac=(-16)^2-4(1)(k)=256-4k\). Two real and distinct roots require \(D>0\), so \(256-4k>0\), which gives \(k<64\). When \(k=64\), the roots are equal, so that option is not correct. Exam tip: Check the sign of the discriminant first when determining the nature of roots.

For this quadratic equation, \(a=1\), \(b=6\), and \(c=k\). Its discriminant is \(D=b^2-4ac=36-4k\). No real roots occur when \(D<0\), so \(36-4k<0\), which gives \(k>9\). At \(k=9\), the equation has one repeated real root, while \(k<9\) gives two real roots. Exam tip: Check the sign of the discriminant first to determine the nature of the roots.

For a quadratic equation to have real roots, its discriminant must satisfy \(D\geq0\). Here, \(a=3\), \(b=-4\), and \(c=p\), so \(D=b^2-4ac=(-4)^2-4(3)(p)=16-12p\). Thus, \(16-12p\geq0\), which gives \(p\leq\frac{4}{3}\). Hence, option A is correct. Option B reverses the required inequality, while \(p<0\) is only a sufficient condition, not the complete condition. Exam tip: For questions about real roots, begin by applying \(D\geq0\).

For a quadratic equation \(ax^2+bx+c=0\) to have equal roots, its discriminant must be zero: \(D=b^2-4ac=0\). Here, \(a=2\), \(b=m\), and \(c=18\), so \(m^2-4(2)(18)=0\), giving \(m^2=144\). Therefore, \(m=\pm12\), or \(m=12\) or \(m=-12\). Option B results from an incorrect calculation of \(4ac\). Exam tip: whenever equal roots are mentioned, immediately use \(D=0\).

For equal roots, the discriminant \(D=b^2-4ac\) must be zero. Here, \(a=k\), \(b=10\), and \(c=5\), so \(10^2-4(k)(5)=0\). Thus, \(100-20k=0\), giving \(k=5\). Exam tip: For equal roots of a quadratic equation, set the discriminant equal to zero.

For the quadratic equation, \\(a=5\\), \\(b=2k\\), and \\(c=2\\). Real roots require the discriminant \\(D=b^2-4ac\\ge0\\). Thus, \\(D=(2k)^2-4(5)(2)=4(k^2-10)\\ge0\\), giving \\(k^2\\ge10\\), or \\(k\\le-\\sqrt{10}\\) or \\(k\\ge\\sqrt{10}\\). In option B, \\(D<0\\), so the roots are not real. Exam tip: For real roots, use \\(D\\ge0\\); equality represents two equal real roots.

A quadratic equation \(ax^2+bx+c=0\) has equal roots when its discriminant \(D=b^2-4ac\) is zero. Here, \(a=1\), \(b=-2(k+4)\), and \(c=25\), so \(D=4(k+4)^2-100=0\). Thus, \((k+4)^2=25\), giving \(k+4=\pm5\). Therefore, \(k=1\) or \(k=-9\), so option A is correct. Exam tip: For equal-root questions, directly apply the condition \(D=0\).

A quadratic equation has no real roots when its discriminant satisfies \(D<0\). Here, \(D=(k+2)^2-4(1)(9)=(k+2)^2-36\). Thus, \((k+2)^2<36\), giving \(-6<k+2<6\), and hence \(-8<k<4\). In option B, \(D>0\), so the equation has two distinct real roots. Exam tip: For ‘no real roots’, always apply the condition \(D<0\).

For a quadratic equation, the discriminant is \\(D=b^2-4ac\\). Here, \\(a=3\\), \\(b=k-2\\), and \\(c=4\\), so \\(D=(k-2)^2-48\\). Two distinct real roots require \\(D>0\\), which gives \\((k-2)^2>48\\). Hence \\(|k-2|>4\\sqrt{3}\\), so \\(k<2-4\\sqrt{3}\\) or \\(k>2+4\\sqrt{3}\\). Therefore, option A is correct. In option B, the discriminant is negative, and at the boundary values the roots are equal. Exam tip: for distinct real roots, always check \\(D>0\\).

For a quadratic equation, \(\Delta>0\) means that the roots are real and distinct. Since \(\Delta=27\) is positive, the roots are distinct and real. Also, 27 is not a perfect square; with rational coefficients, the roots are therefore irrational. Hence, option A is correct. Exam tip: \(\Delta>0\), \(\Delta=0\), and \(\Delta<0\) indicate distinct real, equal real, and non-real roots, respectively.

The discriminant D=b²−4ac links the algebraic roots of a quadratic equation with the graph of its parabola y=ax²+bx+c. Here D=-12, which is negative. A negative discriminant means that the quadratic equation has no real roots, because √D is not real. On the graph, a real root is exactly an x-coordinate where the parabola meets the x-axis. Since there are no real roots, the parabola has no point of intersection with the x-axis. Therefore option A is correct. D=0 would mean that the parabola touches the x-axis once, while D>0 would mean that it cuts the axis at two distinct points. Passing through the origin requires c=0 and is not implied by D=-12.

A parabola intersecting the x-axis at two distinct points means \(D>0\), so the roots are real and distinct. For a quadratic equation with rational coefficients, if \(D\) is not a perfect square, then \(\sqrt{D}\) is irrational; hence, from \(x=\frac{-b\pm\sqrt{D}}{2a}\), both roots are irrational. Therefore, option A is correct. Option B would apply when \(D\) is a perfect square. Exam tip: \(D>0\), \(D=0\), and \(D<0\) indicate distinct real, equal real, and non-real roots, respectively.

For ax²+bx+c=0, two real, rational and distinct roots require D=b²−4ac to be positive and a perfect square. For option A, D=(-17)²−4(1)(72)=289−288=1. Since D=1 is positive and a perfect square, the roots are real, distinct, and rational; in fact, they are (17±1)/2, namely 8 and 9. For option B, D=289−320=-31, so there are no real roots. Option C also has D=-31 and therefore no real roots. In option D, D=(-18)²−4(81)=324−324=0, so the roots are real and equal, not distinct. Hence option A is the only correct choice.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=2\), \(b=-3\), and \(c=5\), so \(D=(-3)^2-4(2)(5)=9-40=-31\). Therefore, option A is correct. Since the discriminant is negative, the equation has no real roots. Exam tip: even when \(b\) is negative, \(b^2\) is positive.

Here, \(a=7\), \(b=-14\), and \(c=7\). Therefore, the discriminant is \(D=b^2-4ac=(-14)^2-4(7)(7)=196-196=0\). For a quadratic equation, \(D=0\) means that the two roots are real and equal. Hence, both the assertion and the reason are correct, and the reason correctly explains the assertion. Exam tip: Calculate \(D=b^2-4ac\) first to determine the nature of the roots.