Nature of Roots

Here (D=(4k)^2-4(4)(9)=16(k^2-9)). For real roots (k^2\geq9), so (k\leq-\frac{3}{2}) or (k\geq\frac{3}{2}).

A quadratic equation \(ax^2+bx+c=0\) has equal roots when its discriminant \(D=b^2-4ac\) is zero. Here, \(a=1\), \(b=-2(k+2)\), and \(c=9\). Thus, \(D=4(k+2)^2-36=0\), giving \((k+2)^2=9\) and hence \(k+2=\pm3\). Therefore, \(k=1\) or \(k=-5\). Exam tip: For equal-root questions, immediately apply the condition \(D=0\).

A quadratic equation has no real roots when its discriminant satisfies \(D<0\). Here, \(a=1, b=k-3, c=4\), so \(D=(k-3)^2-16\). Thus, \((k-3)^2<16\), which gives \(-4<k-3<4\), and hence \(-1<k<7\). At the endpoints in option C, \(D=0\), so the equation has two equal real roots. Exam tip: For ‘no real roots’, always apply the condition \(D<0\).

For the quadratic equation, \\(a=2, b=k+1, c=3\\), so the discriminant is \\(D=b^2-4ac=(k+1)^2-24\\). Two distinct real roots require \\(D>0\\). Hence \\((k+1)^2>24\\), which gives \\(k+1<-2\\sqrt{6}\\) or \\(k+1>2\\sqrt{6}\\). Therefore, \\(k<-1-2\\sqrt{6}\\) or \\(k>-1+2\\sqrt{6}\\). In option B, \\(D<0\\), so the roots are not real. Exam tip: \\(D=0\\) gives equal roots; therefore, use \\(D>0\\) for distinct real roots.

The discriminant \(D=18\) is positive, so the roots are real and distinct. Since 18 is not a perfect square, the roots are irrational. Option B is incorrect because distinct real roots are rational only when the discriminant is a perfect square. Exam tip: \(D>0\) gives distinct real roots, \(D=0\) gives equal roots, and \(D<0\) gives no real roots.

When (D=0), the roots are equal real roots. In graph form this means touching the (x)-axis only once.

The roots of a quadratic equation are the x-coordinates where its graph meets the x-axis. If the parabola remains entirely above the x-axis and does not touch or cross it, there is no point with y = 0. Therefore the equation has no real roots. For a quadratic, the discriminant condition for no real roots is D = b² − 4ac < 0. A tangent parabola touches the x-axis once and corresponds to D = 0, so option B describes a different situation. Crossing the axis twice gives D > 0, which is option C. Also, D = 1 is not implied by the graph and would only be a particular positive perfect-square value. Hence option A is correct.

For a quadratic equation \(ax^2+bx+c=0\), the nature of the roots is determined by the discriminant \(\Delta=b^2-4ac\). In option A, \(a=1, b=-11, c=18\), so \(\Delta=(-11)^2-4(1)(18)=49>0\). Hence, it has two real and distinct roots, namely \(2\) and \(9\). Options B and D have \(\Delta=0\), so they have equal real roots, whereas option C has \(\Delta<0\), so it has no real roots. Exam tip: \(\Delta>0\) always indicates two distinct real roots.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=1\), \(b=-4\), and \(c=6\), so \(D=(-4)^2-4(1)(6)=16-24=-8\). Therefore, option A is correct. As an exam tip, remember that \(D<0\) means the equation has no real roots; 16 is only the value of \(b^2\).

Here, \(a=2\), \(b=-8\), and \(c=8\). Therefore, the discriminant is \(D=b^2-4ac=(-8)^2-4(2)(8)=64-64=0\). For a quadratic equation, \(D=0\) means that the two roots are equal. In fact, the equation can be written as \(2(x-2)^2=0\), giving the repeated root \(x=2\). Exam tip: Calculate the discriminant first to determine the nature of the roots.

Here, \(a=1\), \(b=3\), and \(c=7\). Thus, the discriminant is \(D=b^2-4ac=3^2-4(1)(7)=9-28=-19\). Since \(D<0\), the equation has no real roots, so the assertion is correct. This same discriminant criterion makes the reason correct as well. Exam tip: When \(D<0\), the roots are non-real conjugates.

Here, \(a=6\), \(b=7\), and \(c=-3\). Therefore, the discriminant is \(D=b^2-4ac=7^2-4(6)(-3)=49+72=121\). Since \(D>0\) and 121 is a perfect square, the roots are real, rational, and distinct. Option B is incorrect because irrational roots occur when \(D>0\) but is not a perfect square. Exam tip: First check the sign of the discriminant, then check whether it is a perfect square.

Here, \(a=8\), \(b=-4\), and \(c=3\). Thus, the discriminant is \(D=b^2-4ac=(-4)^2-4(8)(3)=16-96=-80\). Since \(D<0\), the equation has no real roots; its roots are complex. Exam tip: A negative discriminant immediately indicates that a quadratic equation has no real roots.

Here, \(a=5\), \(b=-30\), and \(c=45\). Therefore, the discriminant is \(D=b^2-4ac=(-30)^2-4(5)(45)=900-900=0\). Hence, the roots are real and equal. In fact, the equation can be written as \(5x^2-30x+45=5(x-3)^2=0\), giving the repeated root \(x=3\). Options B and D require \(D>0\), but here \(D=0\). Exam tip: For a quadratic equation, \(D=0\) always indicates two real and equal roots.

For a quadratic equation \\(ax^2+bx+c=0\\), the discriminant is \\(D=b^2-4ac\\). Here, \\(a=1,b=9,c=14\\), so \\(D=9^2-4(1)(14)=81-56=25\\). Since \\(D>0\\) and 25 is a perfect square, the roots are real, rational, and distinct. Therefore, option A is correct. Option B would apply only when \\(D=0\\). Exam tip: Use the sign of \\(D\\) to determine whether the roots are real and distinct, equal, or non-real; then check whether \\(D\\) is a perfect square to identify rational roots.

For a quadratic equation \\(ax^2+bx+c=0\\), the discriminant is \\(D=b^2-4ac\\). Here, \\(a=2, b=1, c=2\\), so \\(D=1^2-4(2)(2)=1-16=-15\\). Since \\(D<0\\), the equation has no real roots. Exam tip: \\(D<0\\) indicates non-real complex roots; \\(D=0\\) indicates equal real roots, while \\(D>0\\) indicates distinct real roots.

Here, \(a=7\), \(b=-9\), and \(c=2\). Therefore, the discriminant is \(D=b^2-4ac=(-9)^2-4(7)(2)=25\). Since \(D\) is positive and a perfect square, the roots are real, rational, and distinct. In fact, the roots are \(1\) and \(\frac{2}{7}\), so option A is correct. Exam tip: a positive perfect-square discriminant indicates rational and distinct roots.

Here, \(a=1, b=-3, c=-1\). Therefore, the discriminant is \(D=b^2-4ac=(-3)^2-4(1)(-1)=13\). Since \(D>0\), the roots are real and distinct; since 13 is not a perfect square, they are also irrational. Hence, option A is correct. Option B is wrong because the roots are not rational. Exam tip: Use \(D>0\) to identify real and distinct roots first, then check whether \(D\) is a perfect square to determine rationality.

For irrational distinct roots, (D>0) and (D) must not be a perfect square. (11) satisfies this condition.

Here, \(D=100>0\), so the two roots are real and distinct. Also, 100 is a perfect square, so \(\sqrt{D}=10\) is rational. Since \(a,b,c\) are rational, the formula \(x=\frac{-b\pm\sqrt{D}}{2a}\) shows that both roots are rational. Therefore, option A is correct. Option D is incorrect because a positive perfect-square discriminant gives rational, not irrational, roots. Exam tip: \(D>0\) indicates distinct real roots, and when \(D\) is a perfect square, the roots are rational.

A quadratic equation \\(ax^2+bx+c=0\\) has equal roots when its discriminant satisfies \\(D=b^2-4ac=0\\). Here, \\(a=1\\), \\(b=-2(k-3)\\), and \\(c=16\\). Therefore, \\(4(k-3)^2-64=0\\), so \\(k-3=\\pm4\\), giving \\(k=7\\) or \\(k=-1\\). Exam tip: For equal-root questions, begin by setting the discriminant equal to zero.

For real roots, ((2k-1)^2-12\geq0) is needed. Hence (2k-1\leq-2\sqrt{3}) or (2k-1\geq2\sqrt{3}).

A quadratic equation \(ax^2+bx+c=0\) has two real and equal roots when its discriminant, \(D=b^2-4ac\), is zero. Here, \(a=2\), \(b=-5\), and \(c=q\). Thus, \((-5)^2-4(2)(q)=0\), giving \(25-8q=0\) and hence \(q=\frac{25}{8}\). Exam tip: For equal roots, immediately apply the condition \(D=0\).

For equal roots, the discriminant of a quadratic equation must be zero. Here, \(a=p\), \(b=-14\), and \(c=7\). Thus, \(b^2-4ac=0\) gives \((-14)^2-4(p)(7)=0\), so \(196-28p=0\) and \(p=7\). Therefore, option A is correct. Exam tip: For equal roots, directly use \(b^2-4ac=0\). Choosing 14 does not make the discriminant zero.

For a quadratic equation to have real roots, its discriminant must satisfy \(D\geq0\). Here \(a=1\), \(b=2(k-1)\), and \(c=k+2\), so \(D=4(k-1)^2-4(k+2)=4(k^2-3k-1)\). Thus, \(k^2-3k-1\geq0\). The zeros of this quadratic are \(\frac{3-\sqrt{13}}{2}\) and \(\frac{3+\sqrt{13}}{2}\). Since its leading coefficient is positive, the inequality holds outside the interval between these zeros; hence option A is correct. Option C results from the calculation error of writing \(k^2-3k-4\) instead. Exam tip: Set the discriminant \(D\geq0\), then solve the resulting quadratic inequality using its zeros and sign pattern.