Nature of Roots

Here (D=4(k+1)^2-4(k+5)). (D\geq0) gives (k^2+k-4\geq0), so solve the resulting inequality carefully.

For the given quadratic equation, \(a=1\), \(b=-10\), and \(c=30\). Thus, the discriminant is \(D=b^2-4ac=(-10)^2-4(1)(30)=100-120=-20\). Since \(D<0\), the equation has no real roots. Option B would be correct only if \(D=0\). Exam tip: the sign of the discriminant directly determines the nature of the roots.

Here, \(a=1\), \(b=-21\), and \(c=110\). Thus, the discriminant is \(D=b^2-4ac=(-21)^2-4(1)(110)=441-440=1\). Since \(D>0\) and 1 is a perfect square, the equation has two distinct real rational roots. Therefore, option A is correct; equal roots occur only when \(D=0\). Exam tip: \(D>0\) indicates distinct real roots, and a perfect-square discriminant indicates that the roots are rational.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=1, b=12, c=37\), so \(D=12^2-4(1)(37)=144-148=-4\). Since \(D<0\), the equation has no real roots. The condition \(D=0\), given in option B, would instead indicate two equal real roots. In an exam, first check the sign of the discriminant to determine the nature of the roots.

Here, \(a=5, b=-14, c=11\). Therefore, the discriminant is \(D=b^2-4ac=(-14)^2-4(5)(11)=196-220=-24\). Since \(D<0\), the equation has no real roots. Option D results from a sign error that gives \(D=24\). Exam tip: \(D<0\) means no real roots, \(D=0\) means equal real roots, and \(D>0\) means distinct real roots.

Here, \(a=1\), \(b=-19\), and \(c=90\). Therefore, the discriminant is \(\Delta=b^2-4ac=(-19)^2-4(1)(90)=361-360=1\). Since \(\Delta\) is positive and a perfect square, the roots are real, rational, and distinct; in fact, they are 9 and 10. Option B would require \(\Delta=0\). Exam tip: \(\Delta>0\) indicates distinct real roots, and a perfect-square discriminant indicates that they are rational.

Here (D=(-5)^2-4(4)(-2)=57). (57) is positive but not a perfect square, so the roots are irrational and distinct.

For the given quadratic equation, \(a=1\), \(b=-2(k-2)\), and \(c=k^2\). Therefore, the discriminant is \(D=b^2-4ac=4(k-2)^2-4k^2=16(1-k)\). No real roots require \(D<0\), so \(16(1-k)<0\), which gives \(k>1\). At \(k=1\), however, \(D=0\), giving two equal real roots. Exam tip: for a quadratic equation, \(D<0\) indicates that the roots are non-real.

Here (D=(-5)^2-4(7)(-3)=109). (109) is positive but not a perfect square, so the roots are irrational and distinct.

For a quadratic equation ax²+bx+c=0, the discriminant D=b²−4ac determines the nature of its roots. Here a=2, b=3 and c=5, so D=3²−4(2)(5)=9−40=−31. Because D is negative, the equation has no real roots; its two roots are complex conjugates. Therefore option C is correct. Equal real roots would require D=0, while two distinct real roots would require D>0. Option D is also incorrect because a non-degenerate quadratic equation does not have exactly one real root under the usual discriminant classification; the negative discriminant gives no real root at all.

A quadratic equation has equal real roots exactly when its discriminant is zero, with the coefficient of x² remaining nonzero. For px²+6x+3=0, identify a=p, b=6, and c=3. Hence D=b²−4ac=6²−4(p)(3)=36−12p. Set D=0: 36−12p=0, so 12p=36 and p=3. Since p=3 is nonzero, the coefficient of x² is nonzero and the expression is genuinely quadratic. Therefore option C is correct. Checking the distractors confirms the result: p=1 gives D=24, p=2 gives D=12, and p=4 gives D=−12. These correspond respectively to distinct real roots for the first two values and non-real roots for p=4, not equal real roots.

The x-intercepts of the parabola y=x²+2x+5 are obtained by solving x²+2x+5=0. Its discriminant is D=b²−4ac=2²−4(1)(5)=4−20=−16. A negative discriminant means the quadratic has no real zeros. Consequently, the graph has no point whose y-coordinate is zero, so it does not meet or cut the x-axis. The assertion is therefore true, and the stated reason is also true and directly explains it. Hence option A is correct. If D were zero, the graph would touch the x-axis once; if D were positive, it would cut the axis at two points. Options B, C, and D incorrectly reject either the assertion, the calculation, or both.

The governing concept is the discriminant condition for coincident roots. Equal real roots occur when D=0. Since D=25−4q², impose the condition 25−4q²=0. Rearranging gives 4q²=25, so q²=25/4. Taking square roots gives q=±√(25/4)=±5/2; both signs must be included because a positive and a negative number have the same square. Thus option A is correct. Option B results from the error q²=25, which ignores the factor 4. For q=±2, D=25−16=9, not zero, so the roots are distinct real roots. For q=0, D=25, also not zero. These checks rule out the remaining choices.

For a quadratic ax² + bx + c = 0, the discriminant is D = b² - 4ac. In this equation, a = 1, b = -2r, and c = r² - 16. Therefore D = (-2r)² - 4(1)(r² - 16) = 4r² - 4r² + 64 = 64. This value is positive for every real r, so the roots are always real and distinct. Moreover, 64 is a perfect square, so the quadratic formula produces rational roots: x = (2r ± 8)/2 = r ± 4. Hence option A is correct. Equal roots would require D = 0, no real roots require D < 0, and irrational roots would require a positive but non-square discriminant.

The governing concept is the discriminant D = b² − 4ac, which determines the nature of quadratic roots. For 2x² − 3√2 x + 7 = 0, we have a = 2, b = −3√2, and c = 7. Thus b² = (−3√2)² = 18, while 4ac = 4 × 2 × 7 = 56. Therefore D = 18 − 56 = −38. Since the discriminant is negative, the equation has no real roots; its two roots are complex conjugates. Hence option A is correct. Option B would require D = 0, and options C and D incorrectly assert positive discriminants and real roots. The irrational coefficient does not change the discriminant rule; it must simply be squared accurately.

For a quadratic equation ax² + bx + c = 0, the discriminant D = b² − 4ac determines the nature of its roots. Here a = 1, b = −2p and c = p² − 25. Therefore, D = (−2p)² − 4(1)(p² − 25) = 4p² − 4p² + 100 = 100. This is positive, so the roots are real and distinct. It is also a perfect square, and the quadratic formula gives x = [2p ± 10]/2 = p ± 5, which are rational whenever p is rational; in the intended school classification, the constant square-root part confirms rational distinct roots. Hence option A is correct. Options B and C would require D = 0 and D < 0 respectively, while D being non-square would lead to irrational roots.

For a quadratic equation ax² + bx + c = 0, the discriminant is D = b² − 4ac. Compare x² − 2(k−2)x + k² = 0 with the standard form. Here a = 1, b = −2(k−2), and c = k². Therefore D = [−2(k−2)]² − 4(1)(k²) = 4(k−2)² − 4k². Expanding gives 4(k² − 4k + 4) − 4k² = 4k² − 16k + 16 − 4k² = 16 − 16k = 16(1−k). Hence option A is correct. Option B results from incorrectly discarding the k-dependent terms. Option C has an incorrect sign and expression, while option D is not obtained from b² − 4ac. Because k is a parameter, the correct discriminant remains an algebraic expression in k.

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 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.

The governing concept is the discriminant D = b² − 4ac. For 5x² − 3√5x + 6 = 0, the coefficients are a = 5, b = −3√5, and c = 6. Hence b² = (−3√5)² = 9 × 5 = 45, while 4ac = 4 × 5 × 6 = 120. Therefore D = 45 − 120 = −75. A negative discriminant means the equation has no real roots; its roots are a complex conjugate pair. D = 0 would describe equal real roots, while positive values would indicate two distinct real roots. Consequently option A is correct. The numerical discriminants quoted in options B, C, and D are also inconsistent with the actual calculation.

The governing concept is the discriminant of a quadratic equation ax² + bx + c = 0. Its value is D = b² − 4ac. Here a = 1, b = −2p and c = p² − 64. Therefore, D = (−2p)² − 4(1)(p² − 64) = 4p² − 4p² + 256 = 256. This is positive for every real value of p, so the roots are real and distinct. Moreover, 256 = 16² is a perfect square, and the coefficients are real, so the roots are rational. In fact, the roots are p + 8 and p − 8. Thus option A is correct. Option B would require D = 0, option C would require D < 0, and option D is wrong because the discriminant is a perfect square.

The governing rule is that a quadratic equation has equal roots exactly when its discriminant is zero. Hence we set the given expression equal to zero: 49 − 16h² = 0. Rearranging gives 16h² = 49, so h² = 49/16. Taking both square roots is essential because both positive and negative values have the same square. Therefore h = ±√(49/16) = ±7/4. Thus option A is correct. Substitution confirms it: for h = 7/4 or −7/4, 16h² = 49 and D = 0. Values ±7 would give D = 49 − 784, while ±4 would give D = 49 − 256; neither produces zero. The value h = 0 gives D = 49, which indicates distinct real roots, not equal roots.