Nature of Roots

Here, \(a=2\), \(b=5\), and \(c=2\). The discriminant is \(D=b^2-4ac=5^2-4(2)(2)=9\). Since \(D>0\), the equation has two real and distinct roots. Exam tip: To determine the nature of roots, first check the sign of the discriminant—\(D>0\) means real and distinct roots.

Here, \(a=9\), \(b=-6\), and \(c=1\). The discriminant is \(D=b^2-4ac=(-6)^2-4(9)(1)=36-36=0\). When \(D=0\), the roots are real and equal; in fact, the repeated root is \(x=\frac{1}{3}\). Exam tip: Use the sign of the discriminant to determine the nature of the roots.

Here, \(a=1\), \(b=1\), and \(c=3\). The discriminant is \(D=b^2-4ac=1^2-4(1)(3)=-11\). Since \(D<0\), the equation has no real roots; its roots are complex. Exam tip: \(D<0\) means no real roots, whereas \(D=0\) gives two equal real roots.

In the standard form \(ax^2+bx+c=0\), we have \(a=6\), \(b=-5\), and \(c=1\). Thus, the discriminant is \(D=b^2-4ac=(-5)^2-4(6)(1)=25-24=1\). Since \(D>0\), the equation also has two real and distinct roots. Exam tip: identify the complete coefficient \(b\) and square it; its negative sign disappears on squaring.

Here, \(a=7\), \(b=2\), and \(c=3\). The discriminant is \(D=b^2-4ac=2^2-4(7)(3)=4-84=-80<0\). Therefore, the equation has no real roots. Exam tip: for a quadratic equation, \(D<0\) means no real roots, while \(D=0\) means two equal real roots.

For distinct real roots (D>0), so ((-2)^2-4n>0) gives (n<1). Use a strict inequality for distinct roots.

For no real roots (D<0), so (16-4p<0) gives (p>4). A negative discriminant gives no real roots.

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=-10, c=r\\), so \\(D=(-10)^2-4(1)(r)=100-4r=0\\), giving \\(r=25\\). Therefore, option A is correct. Exam tip: Whenever equal roots are mentioned, immediately use the condition \\(D=0\\).

For a quadratic equation ax²+bx+c=0, equal real roots occur precisely when the discriminant is zero: Δ=b²−4ac=0. In this equation, a=3, b=k, and c=12. Substitution gives k²−4(3)(12)=0, so k²−144=0 and hence k=±12. Among the listed choices, only 12 appears, so option A is the valid answer. Direct verification is also possible: for k=12, the equation becomes 3x²+12x+12=0, which factors as 3(x+2)²=0 and has the repeated real root x=−2. The value −12 would also work mathematically, but it is not offered. Values 6, 9, and 3 give nonzero discriminants, so their roots are not equal.

A quadratic equation \(ax^2+bx+c=0\) has real and distinct roots when its discriminant satisfies \(D=b^2-4ac>0\). Here, \(a=2\), \(b=3\), and \(c=\lambda\), so \(D=3^2-4(2)(\lambda)=9-8\lambda\). Thus, \(9-8\lambda>0\), which gives \(\lambda<\frac{9}{8}\). When \(\lambda=\frac{9}{8}\), the roots are equal, so option B is not correct. Exam tip: For two distinct real roots, always use the condition \(D>0\).

Here, \(a=1, b=12, c=36\). The discriminant is \(D=b^2-4ac=12^2-4(1)(36)=0\), so the two roots are real and equal. In fact, \(x^2+12x+36=(x+6)^2\), giving both roots as \(x=-6\). Therefore, option C is incorrect because the roots are not distinct. Exam tip: For a quadratic equation, \(D=0\) indicates equal real roots.

Here, \(a=1\), \(b=0\), and \(c=-9\). The discriminant is \(D=b^2-4ac=0^2-4(1)(-9)=36>0\). Therefore, the roots are real and distinct; in fact, they are \(3\) and \(-3\). Equal roots require \(D=0\), so option B is incorrect. Exam tip: \(D>0\) indicates real and distinct roots, \(D=0\) indicates equal roots, and \(D<0\) indicates non-real roots.

Here, \(a=1\), \(b=0\), and \(c=9\). The discriminant is \(D=b^2-4ac=0^2-4(1)(9)=-36<0\), so the equation has no real roots. In fact, its roots are \(\pm 3i\), which are complex. Exam tip: For a quadratic equation, \(D<0\) indicates that there are no real roots.

Here, \(a=4\), \(b=0\), and \(c=-1\). The discriminant is \(D=b^2-4ac=0^2-4(4)(-1)=16>0\), so the equation has two real and distinct roots. In fact, the roots are \(x=\frac{1}{2}\) and \(x=-\frac{1}{2}\). Exam tip: \(D>0\) always indicates two distinct real roots.

Here, \(a=2\), \(b=0\), and \(c=8\). Therefore, the discriminant is \(D=b^2-4ac=0^2-4(2)(8)=-64<0\). When \(D<0\), a quadratic equation has no real roots. Hence, option A is correct; option C would require \(D=0\). Exam tip: Determine the nature of the roots directly from the sign of the discriminant.

For \(x^2+2x+5=0\), \(D=b^2-4ac=2^2-4(1)(5)=-16\). A negative discriminant gives non-real conjugate roots. Option B has equal real roots because its discriminant is zero. Exam tip: check the sign of \(D\) first.

A quadratic equation has two distinct real roots when its discriminant \(D=b^2-4ac\) is positive, that is, \(D>0\). When \(D=0\), the roots are equal, and when \(D<0\), there are no real roots. In an exam, remember the direct correspondence: distinct real roots mean \(D>0\).

For a quadratic equation, the discriminant is \(D=b^2-4ac\). If \(D<0\), then \(\sqrt{D}\) is not real, so the equation has no real roots. When \(D=0\), there are two equal real roots, while \(D>0\) gives two distinct real roots. Exam tip: remember that real roots require \(D\geq0\), whereas no real roots require \(D<0\).

Here, \\(a=1, b=-6, c=9\\). Thus, the discriminant is \\(D=b^2-4ac=(-6)^2-4(1)(9)=0\\). When \\(D=0\\), a quadratic equation has two equal real roots. In fact, \\(x^2-6x+9=(x-3)^2\\), so both roots are 3. Exam tip: remember that \\(D=0\\) indicates equal real roots; option D is incorrect because the discriminant is 0, not 9.

Here, \(a=2\), \(b=-7\), and \(c=3\). The discriminant is \(D=b^2-4ac=(-7)^2-4(2)(3)=49-24=25\). Since \(D>0\), the equation has two real and distinct roots. Option D is incorrect because the sum of the roots is \(-b/a=7/2\), which is positive, so both roots cannot be negative. Exam tip: \(D>0\) indicates real and distinct roots, \(D=0\) indicates real and equal roots, and \(D<0\) indicates no real roots.

Here, \(a=4\), \(b=4\), and \(c=5\). The discriminant is \(D=b^2-4ac=4^2-4(4)(5)=16-80=-64<0\). Hence, the equation has no real roots; its roots are complex. Exam tip: For a quadratic equation, \(D<0\) indicates no real roots.

For the given equation, \(a=1\), \(b=5\), and \(c=6\). Thus, \(\Delta=b^2-4ac=5^2-4(1)(6)=25-24=1\). Since \(\Delta>0\), the equation has two real and distinct roots. Option B would be correct only if \(\Delta=0\). In exams, calculate the discriminant first and then use its sign to determine the nature of the roots.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=3\), \(b=-2\), and \(c=4\), so \(D=(-2)^2-4(3)(4)=4-48=-44\). Since \(D<0\), the discriminant is negative and the equation has no real roots. It is not zero; a zero discriminant occurs only when \(b^2=4ac\). In an exam, first identify \(a,b,c\) and then apply \(D=b^2-4ac\).

Here, \(a=2\), \(b=4\), and \(c=2\). Thus, \(D=b^2-4ac=4^2-4(2)(2)=16-16=0\). Therefore, the discriminant is zero, and the equation has two equal real roots. Choosing positive or negative is incorrect. Exam tip: when \(D=0\), the roots are real and equal.

Here, \(a=1\), \(b=-11\), and \(c=30\). The discriminant is \(D=b^2-4ac=(-11)^2-4(1)(30)=121-120=1\). Since \(D>0\), the roots are real and distinct. In fact, the equation factors as \((x-5)(x-6)=0\), giving roots 5 and 6. Exam tip: \(D>0\) indicates real and distinct roots, \(D=0\) indicates real and equal roots, and \(D<0\) indicates non-real roots.