Nature of Roots

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.

First verify the reason by calculating the discriminant. In x²−6x+11=0, a=1, b=-6, and c=11. Therefore D=b²−4ac=(-6)²−4(1)(11)=36−44=-8. The reason is true. Since D<0, the quadratic equation has no real roots; its roots are a complex conjugate pair. Consequently, the assertion that the roots are real and irrational is false. The word irrational applies to non-real? No: irrational numbers are real numbers that cannot be expressed as a ratio of integers, so these complex roots are not called irrational roots. Thus option A correctly states that the assertion is false while the reason is true. The other choices either incorrectly accept the assertion or reject the correctly calculated discriminant.

Here, \\(a=8, b=2, c=-3\\). Therefore, the discriminant is \\(D=b^2-4ac=2^2-4(8)(-3)=100\\). Since \\(D>0\\), the roots are real and distinct; since \\(100\\) is a perfect square, they are also rational. In fact, the roots are \\(\frac{1}{2}\\) and \\(-\frac{3}{4}\\). Exam tip: If \\(D>0\\) and the discriminant is a perfect square, the roots are real, distinct, and rational.

Here, \(a=7\), \(b=3\), and \(c=2\). Thus, the discriminant is \(D=b^2-4ac=3^2-4(7)(2)=9-56=-47\). Since \(D<0\), the equation has no real roots. Option B would be correct only if \(D=0\), but the discriminant here is negative. Exam tip: To determine the nature of the roots, first calculate the sign of \(D=b^2-4ac\).

Here, \(a=12\), \(b=-12\), and \(c=3\). Therefore, the discriminant is \(D=b^2-4ac=(-12)^2-4(12)(3)=144-144=0\). When \(D=0\), the roots are real and equal. In fact, the equation can be written as \(3(2x-1)^2=0\), giving the repeated root \(x=\frac{1}{2}\). Option B is incorrect because \(D>0\) gives two distinct real roots. Exam tip: For a quadratic equation, \(D=0\) always indicates equal real roots.

The discriminant is \(D=b^2-4ac\). Equal real roots require \(D=0\), so \(b^2=4ac\). If \(D>0\), the roots are distinct real roots. Exam tip: determine the nature of roots from the sign of the discriminant first.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=3, b=1, c=4\), so \(D=1^2-4(3)(4)=1-48=-47\). Since \(D<0\), the equation has no real roots. Exam tip: \(D<0\) indicates no real roots, whereas \(D=0\) indicates two equal real roots.

The discriminant is \\(D=b^2-4ac=(-19)^2-4(10)(6)=121\\). Since \\(D>0\\), the roots are real and distinct. Also, \\(121\\) is a perfect square, so both roots are rational. In fact, the roots are \\(x=\frac{2}{5}\\) and \\(x=\frac{3}{2}\\). Therefore, option A is correct; option D is incorrect because the discriminant is a perfect square. Exam tip: Use \\(D>0\\) for distinct real roots, and check whether \\(D\\) is a perfect square to decide if they are rational.

Here, \(a=2\), \(b=-7\), and \(c=1\). Therefore, the discriminant is \(D=b^2-4ac=(-7)^2-4(2)(1)=41\). Since \(D>0\), the roots are real and distinct; because 41 is not a perfect square, they are also irrational. Option B is incorrect because rational roots in this test require a positive perfect-square discriminant. Exam tip: when \(D>0\) and is not a perfect square, the roots are real, distinct, and irrational.

The direct answer is A: D2=15 gives irrational distinct roots. For a quadratic equation, D>0 gives two real distinct roots. If D is a perfect square, the roots can be rational when the coefficients are rational; if D is not a perfect square, the roots are irrational. Check each value: D1=64 is positive but 64 is a perfect square, so it does not give irrational roots. D2=15 is positive and 15 is not a perfect square, so it gives real irrational distinct roots. D3=0 gives equal real roots. D4=-9 gives no real roots. Therefore option A is correct. The safest exam chain is: first check positive, then check whether the discriminant is a perfect square.

The discriminant \(\Delta=144\) is a positive perfect square. Since \(\Delta>0\), the two roots are real and distinct; because the coefficients are rational and the discriminant is a perfect square, the roots are rational. Hence, option A is correct. Option B is true only when \(\Delta=0\). Exam tip: first use the sign of \(\Delta\) to determine whether the roots are real and distinct, and then check whether it is a perfect square to determine rationality.

For equal roots, the discriminant \(D=b^2-4ac\) must be zero. Here, \(a=1\), \(b=-2(k-5)\), and \(c=36\). Thus, \(D=4(k-5)^2-144=0\), which gives \((k-5)^2=36\). Therefore, \(k-5=\pm6\), so \(k=11\) or \(k=-1\). Exam tip: For equal roots of a quadratic equation, set the discriminant directly equal to zero.

For real roots, ((2k+1)^2-40\geq0) is needed. Hence (2k+1\leq-2\sqrt{10}) or (2k+1\geq2\sqrt{10}).

A quadratic equation has two real and equal roots when its discriminant satisfies \\(D=b^2-4ac=0\\). Here, \\(a=4, b=-9, c=q\\), so \\(D=(-9)^2-4(4)(q)=81-16q\\). Setting this equal to zero gives \\(q=\\frac{81}{16}\\). Exam tip: for equal roots, immediately use the condition \\(D=0\\).

For equal roots, the discriminant must be zero: \(D=b^2-4ac=0\). Here, \(a=p\), \(b=-18\), and \(c=9\), so \((-18)^2-4(p)(9)=0\), giving \(324-36p=0\) and hence \(p=9\). Exam tip: For equal roots, immediately apply the condition \(D=0\).