Nature of Roots

Here, \(a=1\), \(b=-2(2+\sqrt{3})\), and \(c=7+4\sqrt{3}\). Therefore, the discriminant is \(D=b^2-4ac=4(2+\sqrt{3})^2-4(7+4\sqrt{3})=0\). Hence, the two roots are real and equal. In fact, the repeated root is \(x=2+\sqrt{3}\). Option D is incorrect because the root is irrational but not distinct. Exam tip: When \(D=0\), a quadratic equation has two real and equal roots.

The governing test is the discriminant D = b² − 4ac for ax² + bx + c = 0. In this equation, a = 1, b = −2r and c = r² − 49. Thus D = (−2r)² − 4(1)(r² − 49) = 4r² − 4r² + 196 = 196. Since D is positive, the equation has two real and distinct roots. Since 196 = 14² is a perfect square, the roots have the rational form x = [2r ± 14]/2 = r ± 7. Consequently, the correct classification is two real, rational and distinct roots. Equal roots would require D = 0, and no real roots would require D < 0. Option D is unsuitable because the discriminant is a perfect square, not a non-square. Therefore option A is the only correct answer.

The discriminant of a quadratic equation is \(D=b^2-4ac\). Here, \(a=1\), \(b=-2r\), and \(c=r^2+25\), so \(D=(-2r)^2-4(1)(r^2+25)=-100\). Since \(D<0\) for every real value of \(r\), the equation has no real roots. Option B would apply only if \(D=0\), but the discriminant here is always negative. Exam tip: For a quadratic with real coefficients, \(D<0\) means that it has no real roots.

On the x-axis, \(y=0\). Therefore, the intersection equation is \(x^2-2kx+k^2+4=0\), whose discriminant is \(D=(-2k)^2-4(k^2+4)=-16<0\). Hence it has no real roots, so the parabola does not intersect the x-axis. Option D is incorrect because the discriminant remains -16 for every real value of k. Exam tip: Rewriting the equation as \(y=(x-k)^2+4\) immediately shows that the minimum y-value is 4, so the parabola stays 4 units above the x-axis.

The equation \(y=x^2-2(k-1)x+(k-1)^2\) can be rewritten as \(y=(x-(k-1))^2\). Hence its roots are equal and the discriminant is \(D=0\). The vertex is \((k-1,0)\), so for every real value of \(k\), the parabola touches the \(x\)-axis at exactly one point. Option D is incorrect because this tangency is not restricted to \(k=1\). Exam tip: When \(D=0\), a quadratic has equal roots and its parabola touches the \(x\)-axis.

Here, \(a=1, b=-2k, c=k^2-9\). Therefore, the discriminant is \(D=b^2-4ac=(-2k)^2-4(1)(k^2-9)=36\), which is positive for every real value of \(k\). Hence, the equation has two distinct real roots, so the parabola intersects the \(x\)-axis at two distinct points. In fact, the roots are \(x=k+3\) and \(x=k-3\). Option B would require \(D=0\), which is not the case here. Exam tip: \(D>0\) indicates two distinct real intersection points.

For this quadratic, the discriminant is \(D=[-2(a+4)]^2-4(a^2+6a+13)=4(2a+3)\). For \(L\) to be real, \(D\geq0\) is required. Thus, \(4(2a+3)\geq0\), giving \(a\geq-\frac{3}{2}\). In option B, the discriminant is negative, so the roots are not real. Exam tip: use \(D\geq0\) for real roots; use \(D>0\) when two distinct real roots are required.

Here, \(a=1\), \(b=-2p\), and \(c=p^2-7p\). Therefore, the discriminant is \(D=b^2-4ac=4p^2-4(p^2-7p)=28p\). Two real and distinct roots require \(D>0\), so \(28p>0\), which gives \(p>0\). At \(p=0\), \(D=0\) and the roots are equal, while for \(p<0\), the roots are not real. Exam tip: for distinct real roots, always check the condition \(D>0\).

Here \(\Delta=b^2-4ac\). When \(\Delta>0\), the roots are real and distinct; if it is not a perfect square, \(\sqrt{\Delta}\) is irrational. In exams, check the sign of \(\Delta\) first and then whether it is a perfect square.

A quadratic equation has no real roots when its discriminant satisfies \(D<0\). Here, \(D=[-2(a+2)]^2-4(a^2+6a+8)=-8(a+2)\). Thus, \(-8(a+2)<0\), which gives \(a>-2\). When \(a=-2\), \(D=0\), so the equation has two equal real roots; hence it is not correct. Exam tip: For ‘no real roots’, always apply the condition \(D<0\).

The discriminant is \(D=[-2(k+1)]^2-4(k^2+1)=8k\). Distinct real roots require \(D>0\), so \(k>0\). At \(k=0\), the roots are equal. Exam tip: check the sign of the discriminant.

Since ( p\ne 1 ), we have ( (p-1)^2>0 ). Therefore, ( D=-(p-1)^2<0 ). A negative discriminant means that the quadratic equation has no real roots, so option A is correct. Exam tip: ( D=0 ) gives real and equal roots, while ( D>0 ) gives real and distinct roots.

Here, \(a=1\), \(b=-2(1+\sqrt{5})\), and \(c=6+2\sqrt{5}\). Therefore, the discriminant is \(D=b^2-4ac=4(1+\sqrt{5})^2-4(6+2\sqrt{5})=0\), since \((1+\sqrt{5})^2=6+2\sqrt{5}\). Hence, the roots are real and equal; in fact, the repeated root is \(x=1+\sqrt{5}\). Although this root is irrational, the roots are still equal. Exam tip: determine the nature of the roots from the sign of \(D\) before checking whether the root is rational or irrational.

The governing criterion is the discriminant of a quadratic equation. The quadratic formula gives roots x = (−b ± √D)/(2a), with a ≠ 0. Since D > 0, √D is positive and the plus and minus choices produce two distinct real roots. If D were a perfect square, rational coefficients would generally lead to rational roots, but the question states that D is not a perfect square; therefore √D is irrational, making the roots irrational in the standard school-level setting with rational coefficients. D = 0 would instead give equal roots, and D < 0 would give no real roots. Hence option A precisely describes the roots.

Here, \(a=6\), \(b=-4\sqrt{6}\), and \(c=4\). Thus, the discriminant is \(D=b^2-4ac=(-4\sqrt{6})^2-4(6)(4)=96-96=0\). A quadratic equation with \(D=0\) has two real and equal roots. In fact, the repeated root is \(x=\frac{\sqrt{6}}{3}\). In an exam, calculate the discriminant first to determine the nature of the roots.

Here, \(a=2\), \(b=-5\sqrt{2}\), and \(c=8\). Therefore, the discriminant is \(D=b^2-4ac=(-5\sqrt{2})^2-4(2)(8)=50-64=-14\). Since \(D<0\), the equation has no real roots. Option D results from treating \((-5\sqrt{2})^2=50\) as the discriminant and forgetting to subtract \(4ac\). Exam tip: for a quadratic equation, \(D<0\) indicates that the roots are not real.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=3\), \(b=-2\sqrt{21}\), and \(c=7\), so \(D=(-2\sqrt{21})^2-4(3)(7)=84-84=0\). When \(D=0\), the roots are real and equal. In fact, the repeated root is \(x=\frac{-b}{2a}=\frac{\sqrt{21}}{3}\). Therefore, option A is correct; option D is wrong because it describes the roots as distinct. Exam tip: Remember that \(D=0\) indicates two equal real roots.

If the roots are \(\alpha\) and \(\beta\), then \(\alpha\beta=\frac{c}{a}\). A negative product means one root is positive and the other is negative. \(b^2-4ac=0\) indicates equal roots. Exam tip: use the product of roots to check their signs.

The relevant principle is that a quadratic has no real roots exactly when its discriminant is negative. Here a = 1, b = −2(k − 4), and c = k² − 10k + 27. Compute D = [−2(k − 4)]² − 4(k² − 10k + 27) = 4(k − 4)² − 4(k² − 10k + 27). Since (k − 4)² = k² − 8k + 16, the expression inside the difference is k² − 8k + 16 − k² + 10k − 27 = 2k − 11. Thus D = 4(2k − 11). The condition D < 0 gives 2k − 11 < 0, so k < 11/2. Equality produces equal real roots, and larger values produce two distinct real roots. Therefore option A is correct.

The discriminant is \(D=(t+7)^2-4(1)(7t)=(t-7)^2\). Two real and distinct roots require \(D>0\), so \((t-7)^2>0\), which is true exactly when \(t\ne7\). Equivalently, the equation factors as \((x-7)(x-t)=0\); when \(t=7\), the two roots coincide. Exam tip: for distinct real roots of a quadratic, always check the condition \(D>0\).