Nature of Roots

For a quadratic equation ax² + bx + c = 0, the discriminant is D = b² − 4ac. It controls the nature of the roots. When D > 0, the square root of D is positive and non-zero, so the quadratic formula gives two different values: x = (−b + √D)/(2a) and x = (−b − √D)/(2a). Both values are real and unequal. D = 0 would give equal real roots, while D < 0 would give no real roots. Therefore option A is correct.

For a quadratic equation ax² + bx + c = 0, the discriminant is D = b² − 4ac. Comparing 4x² + 4x + 1 = 0 with standard form gives a = 4, b = 4, and c = 1. Substitution gives D = 4² − 4(4)(1) = 16 − 16 = 0. Therefore option A is correct. The value 16 in option C is only b² and ignores the subtraction term. Option B results from an incorrect partial calculation, while option D introduces an unjustified negative sign. Since D = 0, the equation also has two equal real roots.

Use the discriminant criterion for ax² + bx + c = 0: D = b² − 4ac. Here a = 1, b = 6, and c = 9, so D = 6² − 4(1)(9) = 36 − 36 = 0. A zero discriminant means the two roots are real and equal. In fact, the equation can be written as (x + 3)² = 0, showing directly that the repeated root is x = −3. Option A would require D > 0, and option C would require D < 0. A quadratic cannot have three roots, so D is correct only as a distractor.

The governing concept is the discriminant. For ax² + bx + c = 0, equal real roots occur exactly when D = b² − 4ac = 0. In x² − 6x + k = 0, a = 1, b = −6 and c = k. Therefore D = (−6)² − 4(1)(k) = 36 − 4k. Setting this equal to zero gives 36 − 4k = 0, so 4k = 36 and k = 9. Hence option B is correct. The value 6 in option A confuses the coefficient with the required constant, while 12 and 36 result from incomplete or incorrect use of the discriminant condition. When D is zero, the repeated root itself is −b/(2a) = 3, but the question asks for k, not the root.

The discriminant determines the nature of the roots of a quadratic equation ax² + bx + c = 0. Its value is D = b² − 4ac. If D > 0, the equation has two distinct real roots; if D = 0, it has two equal real roots; and if D < 0, it has no real roots. Here D = 49, and 49 is positive. Therefore the equation has two distinct real roots, making option A correct. Although equal roots may sometimes be described informally as one repeated value, that situation requires D = 0, not D = 49. A positive discriminant also means the square root term in the quadratic formula is nonzero, so the two values of x are different.

For a quadratic equation ax² + bx + c = 0, the discriminant is D = b² − 4ac. Its sign determines the nature of the roots: D > 0 gives two distinct real roots, D = 0 gives two equal real roots, and D < 0 gives no real roots. Here D = 64, and 64 is positive. Therefore the equation must have two different real roots, so option A is correct. Option B would apply only when the discriminant is exactly zero. Option C would apply for a negative discriminant, while option D is unsuitable because the positive value already determines the nature clearly.

For ax² + bx + c = 0, the discriminant is D = b² − 4ac. For option A, a = 1, b = 1 and c = 4, so D = 1² − 4(1)(4) = 1 − 16 = −15. Hence option A is correct. Checking the alternatives confirms the uniqueness: option B gives 4² − 4(1)(1) = 12; option C gives 1² − 4(2)(2) = −15? This would also equal −15, creating an ambiguity. Therefore the original item has two correct options, A and C, and cannot pass as written. To make it unambiguous, option C should be changed, for example, to 2x² + x + 1 = 0, whose discriminant is −7.

Let the roots be 3 and r. For x² + ax + 12 = 0, Vieta’s product relation gives 3r = 12, so r = 4. The sum of the roots is 3 + 4 = 7. Vieta’s sum relation says the sum equals −a because the coefficient of x² is 1; hence −a = 7 and a = −7. Therefore option A is correct. It can also be verified directly by substituting x = 3: 9 + 3a + 12 = 0 gives 3a = −21 and a = −7. The other options either use an incorrect sign or fail to satisfy the constant term and the given root simultaneously.

For a quadratic equation ax² + bx + c = 0, the discriminant is D = b² − 4ac. For option A, a = 1, b = 1, and c = 3, so D = 1² − 4(1)(3) = 1 − 12 = −11. Checking the other choices confirms that they do not give the required value: option B gives 9 − 4 = 5, option C gives 1 − 16 = −15, and option D gives 16 − 40 = −24. Thus the equation in option A has discriminant exactly −11. A negative discriminant also indicates that its roots are non-real and distinct as complex conjugates.

The nature of roots can be determined by the discriminant D = b² − 4ac. For 4x² − 12x + 9 = 0, a = 4, b = −12, and c = 9. Thus D = (−12)² − 4(4)(9) = 144 − 144 = 0. A zero discriminant means the quadratic has two equal real roots. Indeed, the expression factors as 4x² − 12x + 9 = (2x − 3)², giving x = 3/2 twice. Therefore option A is correct; D > 0 would indicate distinct real roots and D < 0 would indicate no real roots.

For a quadratic equation ax² + bx + c = 0, the discriminant is D = b² − 4ac. Its sign determines the nature of the roots. If D > 0, the equation has two distinct real roots; if D = 0, it has two equal real roots; and if D < 0, it has no real roots. Here D = 64, and 64 is greater than zero. Therefore the equation has two distinct real roots, making option A correct. A positive discriminant does not mean the roots are equal, nor does it specify that one root is zero; those are unsupported conclusions.

Use the discriminant D = b² − 4ac for the quadratic 9x² − 24x + 16 = 0. Here a = 9, b = −24, and c = 16, so D = (−24)² − 4(9)(16) = 576 − 576 = 0. Since the discriminant is zero, the equation has two equal real roots. The same result is visible from factorisation: 9x² − 24x + 16 = (3x − 4)². Thus both roots are x = 4/3, counted twice, and option A is correct. The other choices conflict with the zero-discriminant criterion.

For ax² + bx + c = 0, the discriminant D = b² − 4ac determines the nature of the roots. Here a = 25, b = −30, and c = 9, so D = (−30)² − 4(25)(9) = 900 − 900 = 0. A zero discriminant means the quadratic has two equal real roots. Indeed, the expression factors as 25x² − 30x + 9 = (5x − 3)², giving x = 3/5 twice. Therefore option A is correct. Distinct real roots require D > 0, no real roots require D < 0, and zero is not a root because the constant term is nonzero.

For ax^2+bx+c=0, where a is nonzero, the discriminant is Δ=b^2-4ac. The quadratic formula gives roots (-b±√Δ)/(2a). The roots are equal precisely when the plus and minus expressions coincide, which requires √Δ=0 and therefore Δ=0. Hence b^2-4ac=0, or b^2=4ac, so option A is correct. If b^2>4ac, the discriminant is positive and the roots are distinct real numbers. If b^2<4ac, the discriminant is negative and the roots are non-real conjugates. The relation a+b+c=0 instead only indicates that x=1 is a root; it does not generally imply equal roots.

The governing concept is the discriminant criterion for a quadratic. For ax^2+bx+c=0, roots are non-real when D=b^2−4ac<0. Here a=5, b=−4 and c=λ, so D=(−4)^2−4(5)(λ)=16−20λ. Requiring non-real roots gives 16−20λ<0. Therefore −20λ<−16, and division by the negative number −20 reverses the inequality: λ>16/20=4/5. Hence option C is correct. Equality λ=4/5 gives one repeated real root, while λ<4/5 gives two distinct real roots; this rules out options A, B and D.

For a quadratic equation ax² + bx + c = 0, the discriminant is D = b² − 4ac. The discriminant determines the nature and number of real roots. When D > 0, its square root is a positive nonzero real number, so the quadratic formula x = (−b ± √D)/(2a) produces two different real values because the plus and minus signs give different results. Therefore the equation has two distinct real roots, making option A correct. If D = 0, the two values coincide and there is one repeated real root, so option B would apply. If D < 0, there are no real roots, so option C would apply. Option D is incomplete because a positive discriminant never gives just one root.

The governing concept is the discriminant D = b² − 4ac, which determines the nature of the roots of ax² + bx + c = 0. If D is positive, the equation has two distinct real roots; if D is zero, it has two equal real roots; and if D is negative, the square root of D is not a real number. Here D = −4, so √D = √(−4) is not real. Consequently, the quadratic equation has no real roots, although it would have complex roots if complex numbers were being considered. Therefore option A is correct. Options B and C apply to D = 0 and D > 0 respectively, while option D cannot be concluded from the discriminant alone.

The governing concept is the non-negative-square property: for every real number x, x² is greater than or equal to zero. Rearranging the equation x² + 25 = 0 gives x² = −25. A real square cannot equal a negative number, so no real value of x can satisfy this equation. Equivalently, the discriminant for x² + 25 = 0 is 0² − 4(1)(25) = −100, which is negative and therefore confirms the absence of real roots. Hence option A is correct. Although 5 and −5 are related to the square of 25, they solve x² − 25 = 0, not x² + 25 = 0. The other options also contradict the equation.

The governing concept is the discriminant of a quadratic equation ax² + bx + c = 0, defined as D = b² − 4ac. Its sign determines the nature of real roots: D > 0 gives two distinct real roots, D = 0 gives two equal real roots, and D < 0 gives no real roots. Since D = −9 is negative, the expression under the square root in the quadratic formula, √D, would be √(−9), which is not a real number. Therefore option A is correct. A negative discriminant does not imply that a root is zero; a zero root would require c = 0. Options B and C apply to D = 0 and D > 0 respectively, so they do not fit this equation.

The governing property is that the square of every real number is non-negative: x² ≥ 0. Rearranging the equation gives x² = −36. No real number can have a square equal to a negative number, so the equation has no real solution. Equivalently, for x² + 36 = 0, a = 1, b = 0 and c = 36, so the discriminant is D = 0² − 4(1)(36) = −144, which is negative and therefore confirms that there are no real roots. Thus option A is correct. The pair 6 and −6 would solve x² − 36 = 0, not x² + 36 = 0. The number 36 is a constant, not a root, and x = 0 gives 36 rather than zero in the original equation.