Nature of Roots

For a quadratic equation, the discriminant is \(D=b^2-4ac\). When \(D<0\), the roots are not real; they are complex. Since \(D=-1<0\), the equation has no real roots. Exam tip: \(D>0\) gives two real and distinct roots, \(D=0\) gives two real and equal roots, and \(D<0\) gives no real roots.

The discriminant \(D=b^2-4ac\) determines the nature of the roots. When \(D=0\), both roots are real and equal; the repeated root is \(x=-\frac{b}{2a}\). Therefore, option B is incorrect because distinct real roots occur when \(D>0\). Exam tip: remember \(D=0\) as ‘equal real roots’.

Here, \(a=1\), \(b=-12\), and \(c=36\). Therefore, the discriminant is \(D=b^2-4ac=(-12)^2-4(1)(36)=144-144=0\). When \(D=0\), the roots are real and equal. In fact, \(x^2-12x+36=(x-6)^2\), so both roots are \(6\). Exam tip: Find the discriminant first to determine the nature of the roots.

Here, \(a=1, b=-6, c=13\). Therefore, the discriminant is \(\Delta=b^2-4ac=(-6)^2-4(1)(13)=36-52=-16<0\). A negative discriminant means that the quadratic equation has no real roots, so option A is correct. Option C would be correct only if \(\Delta=0\). Exam tip: \(\Delta<0\), \(\Delta=0\), and \(\Delta>0\) indicate no real roots, equal roots, and distinct real roots, respectively.

For \(x^2+px+4=0\), we have \(a=1\), \(b=p\), and \(c=4\). Real and distinct roots require the discriminant \(D=b^2-4ac\) to be positive. Thus, \(D=p^2-16>0\), which gives \(p^2>16\). If \(p^2=16\), the roots are equal, so option B is incorrect. Exam tip: remember that distinct real roots require \(D>0\).

For a quadratic equation \(ax^2+bx+c=0\) to have real and equal roots, its discriminant must be zero: \(D=b^2-4ac=0\). Here, \(a=3\), \(b=m\), and \(c=12\), so \(m^2-4(3)(12)=0\), giving \(m^2=144\). Exam tip: For equal roots, immediately set the discriminant equal to zero.

For a quadratic equation written in standard form ax^2+bx+c=0, where a≠0, the discriminant is defined as D=b^2−4ac. It is the expression under the square root in the quadratic formula x=(−b±√(b^2−4ac))/(2a). The value of D determines the nature of the roots: D>0 gives two real distinct roots, D=0 gives two equal real roots, and D<0 gives non-real conjugate roots. Therefore option A is correct. Options B and C incorrectly interchange the roles of the coefficients, while option D omits the square on b. Correctly identifying a, b and c before substitution prevents these common errors.

A positive perfect square (D) gives real, rational and distinct roots. Check both the sign and perfect square condition of (D).

A positive non-perfect-square (D) gives real, irrational and distinct roots. Only (D>0) tells distinctness.

For a quadratic equation \\(ax^2+bx+c=0\\), the discriminant is \\(D=b^2-4ac\\). Here, \\(a=1, b=-3,c=2\\), so \\(D=(-3)^2-4(1)(2)=9-8=1\\). Therefore, the correct answer is 1. The value 5 results from incorrectly adding instead of subtracting \\(4ac\\). In an exam, first identify \\(a,b,c\\) and then substitute them carefully.

Here \(a=1\), \(b=-3\), and \(c=2\), so the discriminant is \(D=b^2-4ac=(-3)^2-4(1)(2)=1\). Since \(D>0\) and is a perfect square, the roots are real, rational, and distinct. In fact, \(x^2-3x+2=(x-1)(x-2)\), giving the roots \(1\) and \(2\). Exam tip: \(D>0\) indicates distinct roots, and if \(D\) is a perfect square, those roots are rational.

For the given equation, \(a=1\), \(b=-4\), and \(c=1\). Hence, \(D=b^2-4ac=(-4)^2-4(1)(1)=16-4=12\). Therefore, the correct answer is 12. Since \(D>0\), the roots are real and distinct; because 12 is not a perfect square, they are also irrational. Exam tip: Check the sign of the discriminant first to determine the nature of the roots.

For a quadratic equation ax² + bx + c = 0, the discriminant D = b² − 4ac determines the nature of its roots. Here a = 1, b = −4 and c = 1, so D = (−4)² − 4(1)(1) = 16 − 4 = 12. Since D is positive, the equation has two real and distinct roots. Also, 12 is not a perfect square. Using the quadratic formula, the roots are (4 ± √12)/2 = 2 ± √3, and √3 is irrational. Therefore the roots are real, irrational and distinct, making option A correct. Option B is wrong because the roots are not rational; options C and D contradict the positive discriminant.

Here, \(a=2\), \(b=4\), and \(c=2\). Using the discriminant formula \(D=b^2-4ac\), we get \(D=(4)^2-4(2)(2)=16-16=0\). Therefore, the correct answer is 0, and the equation has equal roots. Option 4 is only the value of \(b^2\), not the complete discriminant. In an exam, identify \(a,b,c\) first and substitute them carefully into the formula.

Here, \(a=2\), \(b=4\), and \(c=2\). The discriminant is \(D=b^2-4ac=4^2-4(2)(2)=16-16=0\). When \(D=0\), the roots are real and equal; in fact, both roots are \(-1\). Therefore, option A is correct. Exam tip: remember that \(D=0\) indicates real and equal roots.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=1, b=3, c=7\), so \(D=3^2-4(1)(7)=9-28=-19\). Therefore, option A is correct. Exam tip: a negative discriminant means that the equation has no real roots; 19 results from missing the negative sign.

Here, \(a=1, b=3, c=7\). The discriminant is \(D=b^2-4ac=3^2-4(1)(7)=9-28=-19<0\). Therefore, the equation has no real roots, so the correct answer is 0. Exam tip: For a quadratic equation, \(D<0\) means there are zero real roots.

For a quadratic equation \\(ax^2+bx+c=0\\\), the discriminant is given by \\(D=b^2-4ac\\\). Here, \\(a=3, b=-5, c=2\\\), so \\(D=(-5)^2-4(3)(2)=25-24=1\\\). Since \\(D>0\\\), the equation has two distinct real roots. In an exam, remember to include the negative sign of \\(b\\\) before squaring it.

Here, \(a=3, b=-5, c=2\). The discriminant is \(D=b^2-4ac=(-5)^2-4(3)(2)=25-24=1\). Since \(D\) is positive and a perfect square, the two roots are real, rational, and distinct. In fact, the roots are \(x=1\) and \(x=\frac{2}{3}\). Exam tip: \(D>0\) indicates distinct roots, while a perfect-square discriminant makes them rational.

Here, \(a=2\), \(b=-1\), and \(c=4\). Thus, the discriminant is \(D=b^2-4ac=(-1)^2-4(2)(4)=1-32=-31\). Since \(D<0\), its sign is negative, and the equation has no real roots. In an exam, check the sign of \(b\) carefully before substituting in \(b^2-4ac\).

Here, \(a=2\), \(b=-1\), and \(c=4\). The discriminant is \(D=b^2-4ac=(-1)^2-4(2)(4)=1-32=-31\). Since \(D<0\), the equation has no real roots, so option A is correct. Exam tip: Check the sign of the discriminant first to determine the nature of the roots.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=1\), \(b=-6\), and \(c=8\), so \(D=(-6)^2-4(1)(8)=36-32=4\). Therefore, the correct answer is 4. A discriminant of 0 would indicate equal roots, which is not the case here. Exam tip: identify the signs of \(a\), \(b\), and \(c\) carefully before substituting.

Here, \(a=1, b=-6, c=8\). The discriminant is \(D=b^2-4ac=(-6)^2-4(1)(8)=36-32=4\). Since \(D>0\), the roots are real and distinct; since \(D=4\) is a perfect square, they are also rational. In fact, the roots are \(2\) and \(4\). Therefore, option A is correct. Exam tip: If \(D>0\) and is a perfect square, the roots are real, rational and distinct.

For a quadratic equation 8ax^2+bx+c=0\u00029, the discriminant is 8D=b^2-4ac\u00029. Here, 8a=1, b=-2, c=-1\u00029, so 8D=(-2)^2-4(1)(-1)=4+4=8\u00029. Since 8c=-1\u00029, subtracting 8-4ac\u00029 results in adding 4; therefore, 4 is not the discriminant. Exam tip: identify the signs of 8a,b,c\u00029 before substitution.

Here, \(a=1, b=-2, c=-1\). Therefore, the discriminant is \(D=b^2-4ac=(-2)^2-4(1)(-1)=8\). Since \(D>0\), the roots are real and distinct. Also, 8 is not a perfect square, so the roots are irrational. Exam tip: If \(D>0\) and \(D\) is not a perfect square, the roots are real, irrational, and distinct.