Nature of Roots

The governing principle is that a polynomial of degree n can have at most n roots. Because the coefficient of x² is 2, which is non-zero, the equation 2x² + 7x + 3 = 0 is genuinely quadratic and has degree 2. Therefore, it cannot have more than two real solutions. For this particular equation, the discriminant confirms that the maximum is attained: D = b² − 4ac = 7² − 4(2)(3) = 49 − 24 = 25, which is positive, so there are two distinct real roots. Option A is not the maximum, while options C and D exceed the degree of the equation. Hence option B is correct.

For a quadratic equation ax² + bx + c = 0, the discriminant is D = b² − 4ac. It determines the nature of the roots because the quadratic formula is x = (−b ± √D)/(2a). If D > 0, then √D is real and non-zero. The plus and minus signs consequently produce two different real values of x, so the equation has two distinct real roots. If D = 0, the two values coincide and the roots are equal. If D < 0, √D is not real, so there are no real roots. Therefore option A is correct; the other choices correspond to different discriminant conditions or an incomplete description.

For a quadratic equation ax² + bx + c = 0, the discriminant is calculated using D = b² − 4ac. Comparing 4x² + 4x + 1 = 0 with the standard form gives a = 4, b = 4, and c = 1. Substituting these values, D = 4² − 4(4)(1) = 16 − 16 = 0. Therefore option A is correct. Option C represents only b² and fails to subtract 4ac. Option B results from an incorrect partial calculation, and option D introduces a negative sign that is not produced by the formula. Since D = 0, the equation also has two equal real roots, which confirms the calculation.

Answer: B, two equal real roots. For ax² + bx + c = 0, the discriminant is D = b² − 4ac. Here a = 1, b = 6, and c = 9, so D = 6² − 4(1)(9) = 36 − 36 = 0. The criterion is: D > 0 gives two distinct real roots, D = 0 gives two equal real roots, and D < 0 gives no real roots in the real number system. We can verify the result by factorising: x² + 6x + 9 = (x + 3)². Thus (x + 3)² = 0 gives x = −3 twice, so the repeated root is real and equal. A would require a positive discriminant. C would require a negative discriminant. D is impossible because a quadratic has at most two roots. Memory cue: zero discriminant means the parabola touches the x-axis once, representing one repeated real root.

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.

The governing concept is the discriminant of a quadratic equation. For ax² + bx + c = 0, the discriminant is D = b² − 4ac, and 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, which is positive, so the equation has two distinct real roots. Therefore option A is correct. Option B would be correct only for D = 0, not for 64. Option C applies when the discriminant is negative. Option D is also incorrect because the positive value of the discriminant gives a definite conclusion without requiring the actual coefficients or solving the equation.

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. Therefore option A exactly matches the required discriminant. The remaining choices can be checked to rule them out: option B gives D = 3² − 4(1)(1) = 9 − 4 = 5; option C gives D = 1² − 4(2)(2) = 1 − 16 = −15; and option D gives D = (−4)² − 4(1)(10) = 16 − 40 = −24. Thus only option A has discriminant −11. Its negative discriminant also indicates that its roots are non-real complex conjugates.

The nature of roots is governed by the discriminant D = b² − 4ac for ax² + bx + c = 0. Here a = 4, b = −12, and c = 9. Therefore D = (−12)² − 4(4)(9) = 144 − 144 = 0. A zero discriminant means that the quadratic has two equal real roots. This can also be verified by factoring: 4x² − 12x + 9 = (2x − 3)², so both roots are x = 3/2. Hence option A is correct. Distinct real roots require D > 0, while no real roots occur when D < 0. The equation also does not have a zero root because its constant term is non-zero.

Use the discriminant criterion D = b² − 4ac. For 9x² − 24x + 16 = 0, the coefficients are a = 9, b = −24, and c = 16. Thus D = (−24)² − 4(9)(16) = 576 − 576 = 0. Since D equals zero, the equation has two equal real roots. This is confirmed by the factorisation 9x² − 24x + 16 = (3x − 4)², which gives x = 4/3 twice. Hence option A is correct. Distinct real roots would require a positive discriminant, and no real roots would require a negative one. Because the constant term is 16 rather than zero, x = 0 is not a root, so option D is also impossible.

For a quadratic equation ax² + bx + c = 0, the discriminant D = b² − 4ac determines the nature of its roots. Here a = 25, b = −30, and c = 9. Therefore D = (−30)² − 4(25)(9) = 900 − 900 = 0. When the discriminant is zero, the equation has two real and equal roots. This can also be confirmed by factoring: 25x² − 30x + 9 = (5x − 3)², so the repeated root is x = 3/5. Hence option A is correct. Distinct real roots require D > 0, while no real roots occur when D < 0. Option D is also impossible because the constant term is 9, not zero, so x = 0 cannot be a root.

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 equal roots. For a quadratic ax² + bx + c = 0, equal real roots occur exactly when the discriminant D = b² − 4ac is zero. In x² − 12x + k = 0, the coefficients are a = 1, b = −12, and c = k. Thus D = (−12)² − 4(1)(k) = 144 − 4k. Setting D = 0 gives 144 − 4k = 0, so 4k = 144 and k = 36. Therefore option A is correct. Option B is merely the magnitude of the x-coefficient, option C results from an incomplete division, and option D is 12² without subtracting 4ac; none satisfies the discriminant condition.

The roots of a quadratic equation are \\(\frac{-b\pm\sqrt{D}}{2a}\\). When \\(D>0\\), \\(\sqrt{D}\\) is a positive real number, so the plus and minus forms give two real and distinct roots. For comparison, \\(D=0\\) gives equal roots, while \\(D<0\\) gives no real roots. In an exam, first calculate or determine the sign of the discriminant \\(D=b^2-4ac\\).

Here (D=(-5)^2-4(1)(6)=1), so the roots are real and distinct. A perfect-square (D) gives rational roots.

Here, \(a=1, b=4, c=4\). Therefore, the discriminant is \(D=b^2-4ac=4^2-4(1)(4)=0\). When \(D=0\), a quadratic equation has two real and equal roots. In fact, \(x^2+4x+4=(x+2)^2\), so both roots are \(x=-2\). Exam tip: To determine the nature of roots, first calculate \(D=b^2-4ac\).

Here, \(a=2\), \(b=1\), and \(c=3\). Therefore, the discriminant is \(D=b^2-4ac=1^2-4(2)(3)=-23\). Since \(D<0\), the quadratic equation has no real roots. Options B and C are incorrect because they correspond to \(D=0\) and \(D>0\), respectively. Exam tip: To determine the nature of the roots, first check the sign of the discriminant.

For a quadratic equation \(ax^2+bx+c=0\) to have equal roots, its discriminant must be zero: \(D=b^2-4ac=0\). Here, \(a=1\), \(b=-4\), and \(c=k\). Thus, \((-4)^2-4(1)(k)=0\), giving \(16-4k=0\) and hence \(k=4\). Therefore, option A is correct. Exam tip: for equal roots, immediately use \(D=0\).

For equal roots, the discriminant \(D=b^2-4ac\) must be zero. Here, \(a=1\), \(b=k\), and \(c=9\), so \(D=k^2-36=0\). Thus, \(k^2=36\), giving \(k=6\) or \(k=-6\). Therefore, option A is correct. Exam tip: For equal roots of a quadratic equation, set \(D=0\) and remember to consider both positive and negative square roots.

For a quadratic equation \(ax^2+bx+c=0\) to have two real and distinct roots, its discriminant must satisfy \(\Delta=b^2-4ac>0\). Here, \(a=1, b=-6, c=k\), so \(\Delta=(-6)^2-4(1)(k)=36-4k\). Thus, \(36-4k>0\), which gives \(k<9\). When \(k=9\), the roots are equal, and when \(k>9\), the roots are not real. Exam tip: for distinct real roots, always require \(\Delta>0\).

A quadratic equation has no real roots when its discriminant satisfies \(D<0\). Here, \(D=b^2-4ac=2^2-4(1)(k)=4-4k\). Thus, \(4-4k<0\), which gives \(k>1\). If \(k=1\), then \(D=0\), so the equation has two equal real roots; hence option B is incorrect. Exam tip: Check the sign of the discriminant first to determine the nature of the roots.

The equation \(x^2-8x+16=0\) can be written as \((x-4)^2=0\). Thus, \(x-4=0\), so both equal roots are \(4\). The option \(-4\) is a common sign-error distractor. Exam tip: for equal roots, use \(x=-\frac{b}{2a}\) or check that the discriminant \(D=b^2-4ac=0\).

Here (D=9-4=5), which is positive but not a perfect square. So roots are real distinct and irrational.

For the given quadratic equation, \(a=2\), \(b=-5\), and \(c=2\). Therefore, the discriminant is \(D=b^2-4ac=(-5)^2-4(2)(2)=25-16=9\). Since \(D\) is positive and a perfect square, the roots are real, rational, and distinct. Option B is incorrect because a positive discriminant gives irrational roots only when it is not a perfect square. Exam tip: \(D>0\) indicates distinct real roots, while a perfect-square discriminant indicates that the roots are rational.

For the equation \(3x^2-12=0\), \(a=3\), \(b=0\), and \(c=-12\). Thus, the discriminant is \(D=b^2-4ac=0-4(3)(-12)=144\). Since \(D>0\), the roots are real and distinct; because \(\sqrt{D}=12\) is an integer, both roots are rational. In fact, the roots are \(x=2\) and \(x=-2\). Exam tip: If \(D>0\) and is a perfect square, the roots are real, rational, and distinct.