Nature of Roots

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.

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 discriminant of \(ax^2+bx+c=0\) is \(D=b^2-4ac\). It determines the nature of the roots: \(D>0\) gives two distinct real roots, \(D=0\) gives equal real roots, and \(D<0\) gives no real roots. Therefore, option A is correct; the other expressions are not the discriminant formula. Exam tip: identify the coefficients \(a,b,c\) correctly before substituting their values.

For a quadratic equation
(ax^2+bx+c=0
), the discriminant is
(D=b^2-4ac
). When
(D>0
), the term
\(\sqrt{D}\) is real and non-zero, so the two roots
\((-b+\sqrt{D})/(2a)\) and
\((-b-\sqrt{D})/(2a)\) are real and distinct. Option B applies when
(D=0
), while
(D<0
) gives no real roots. Exam tip: determine the nature of roots by checking the sign of the discriminant first.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant \(D=b^2-4ac\) determines the nature of the roots. When \(D=0\), the term \(\sqrt{D}\) is zero, so both values from \(x=\frac{-b\pm\sqrt{D}}{2a}\) are identical; hence the roots are real and equal. Option B applies when \(D>0\), and the roots are not necessarily positive. Exam tip: remember \(D=0\) as “equal real roots.”

Here, \(a=1, b=-5, c=6\). The discriminant is \(D=b^2-4ac=(-5)^2-4(1)(6)=25-24=1\). Since \(D>0\), the quadratic equation has two real and distinct roots; in fact, the roots are \(2\) and \(3\). Remember: \(D>0\) gives real and distinct roots, \(D=0\) gives equal roots, and \(D<0\) gives no real roots.

Here, \(a=1\), \(b=-4\), and \(c=4\). The discriminant is \(D=b^2-4ac=(-4)^2-4(1)(4)=0\). Therefore, the roots are real and equal; in fact, the equation is \((x-2)^2=0\), so both roots are \(2\). Exam tip: when \(D=0\), a quadratic equation has real and equal roots.

Here, \(a=1\), \(b=2\), and \(c=5\). The discriminant is \(D=b^2-4ac=2^2-4(1)(5)=4-20=-16\). Since \(D<0\), the equation has no real roots, so option A is correct. Exam tip: For a quadratic equation, \(D<0\) indicates non-real complex roots; equal real roots occur only when \(D=0\).

For a quadratic equation \\(ax^2+bx+c=0\\), the discriminant is \\(\Delta=b^2-4ac\\). Here, \\(a=2, b=-3, c=1\\), so \\(\Delta=(-3)^2-4(2)(1)=9-8=1\\). Therefore, the correct answer is 1. Exam tip: Identify the signs of \\(a\\), \\(b\\), and \\(c\\) carefully before substituting them.

Here, \(a=3, b=6, c=3\). The discriminant is \(D=b^2-4ac=6^2-4(3)(3)=0\). When \(D=0\), a quadratic equation has two real and equal roots. In fact, \(3x^2+6x+3=3(x+1)^2\), so both roots are \(-1\). Exam tip: remember that \(D=0\) indicates equal real roots.

Here, \(a=5\), \(b=1\), and \(c=1\). The discriminant is \(D=b^2-4ac=1^2-4(5)(1)=-19<0\). Therefore, the equation has no real roots. Option B is incorrect because real and distinct roots occur only when \(D>0\). Exam tip: use the sign of the discriminant—\(D<0\) means that the quadratic has no real roots.

A quadratic equation \(ax^2+bx+c=0\) has equal real roots when its discriminant \(D=b^2-4ac\) is zero. Here, \(a=1\), \(b=6\), and \(c=k\), so \(D=6^2-4(1)(k)=0\), giving \(36-4k=0\) and hence \(k=9\). Therefore, 9 is correct. Exam tip: For equal roots, immediately use the condition \(D=0\).

For a quadratic equation \\(ax^2+bx+c=0\\), equal real roots occur when the discriminant \\(D=b^2-4ac\\) is zero. Here, \\(a=1, b=-8, c=m\\), so \\(D=(-8)^2-4(1)(m)=0\\), giving \\(64-4m=0\\) and hence \\(m=16\\). Therefore, 16 is correct. Exam tip: For equal roots, set the discriminant directly to zero.

For a quadratic equation \(ax^2+bx+c=0\) to have equal real roots, its discriminant must be zero: \(D=b^2-4ac=0\). Here \(a=1\), \(b=k\), and \(c=16\), so \(D=k^2-64=0\), giving \(k=\pm 8\). Since only 8 appears among the options, option A is correct. Exam tip: set the discriminant equal to zero whenever equal roots are required.

For a quadratic equation ax² + bx + c = 0, the discriminant is D = b² − 4ac. For option A, D = (−7)² − 4(1)(10) = 9 > 0, so its roots are real and distinct. In options B and D, D = 0, giving real and equal roots, while option C has D = 16 − 32 = −16, so it has no real roots. Exam tip: D > 0 indicates two real and distinct roots.

A quadratic equation \(ax^2+bx+c=0\) has two real and equal roots when its discriminant \(D=b^2-4ac\) is zero. For option A, \(D=10^2-4(1)(25)=100-100=0\). Also, it is \((x+5)^2=0\), so both roots are \(-5\). In option B, the discriminant is positive, so it has two distinct real roots. Exam tip: check \(b^2-4ac=0\) for equal real roots.

For a quadratic equation ax² + bx + c = 0, the discriminant is D = b² − 4ac. In option A, D = 3² − 4(1)(5) = −11 < 0, so the equation has no real roots. Options B and D have D > 0 and therefore two distinct real roots, while option C has D = 0 and therefore two equal real roots. Exam tip: check the sign of the discriminant first to determine the nature of the roots.

Here, \(a=4\), \(b=-12\), and \(c=9\). The discriminant is \(D=b^2-4ac=(-12)^2-4(4)(9)=144-144=0\). Therefore, the roots are real and equal. In fact, \(4x^2-12x+9=(2x-3)^2\), so both roots are \(x=\frac{3}{2}\). Exam tip: When \(D=0\), the roots are always real and equal.

For a quadratic equation, the discriminant is \(D=b^2-4ac\). When \(D>0\), the two roots are real and distinct. Since \(D=25>0\), option A is correct. Equal roots occur when \(D=0\), while non-real roots occur when \(D<0\). Exam tip: Check the sign of the discriminant first to determine the nature of the roots.