Nature of Roots

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.

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.