Nature of Roots

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.

For the quadratic equation, \(a=1\), \(b=-6\), and \(c=10\). Thus, the discriminant is \(D=b^2-4ac=(-6)^2-4(1)(10)=36-40=-4\). Since \(D<0\), the equation has no real roots, so \(s\) cannot have a real value in the area problem. Exam tip: \(D<0\), \(D=0\), and \(D>0\) indicate no real roots, equal real roots, and distinct real roots, respectively.

For the given equation, \(a=1\), \(b=-15\), and \(c=56\). Therefore, the discriminant is \(D=b^2-4ac=(-15)^2-4(1)(56)=225-224=1\). Since \(D>0\) and 1 is a perfect square, the roots are real, rational, and distinct. Exam tip: a positive perfect-square discriminant gives two distinct rational roots; merely having \(D>0\) does not guarantee rational roots.

For a quadratic equation \\(ax^2+bx+c=0\\), the discriminant is \\(D=b^2-4ac\\). Here, \\(a=1,b=8,c=18\\), so \\(D=8^2-4(1)(18)=64-72=-8\\). Since \\(D<0\\), the equation has no real roots. Option B would be correct only if \\(D=0\\). Exam tip: Always check the sign of the discriminant before identifying the nature of the roots.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(D=(-12)^2-4(4)(11)=144-176=-32\). Since \(D<0\), the equation has no real roots. Option A is incorrect because a negative discriminant does not produce real irrational roots; it produces non-real complex roots. Exam tip: remember that \(D<0\) means no real roots.

Here, \(a=1, b=-13, c=40\). Therefore, the discriminant is \(D=b^2-4ac=(-13)^2-4(1)(40)=169-160=9\). Since \(D>0\), the roots are real and distinct; because \(D=9\) is a perfect square, they are also rational. In fact, the roots are \(5\) and \(8\). Exam tip: If \(D>0\) and is a perfect square, the roots are real, distinct, and rational.

Here (D=(-4)^2-4(3)(-1)=28). (28) is positive but not a perfect square, so the roots are irrational and distinct.