Nature of Roots

Here, \(A=1\), \(B=-2(a+4)\), and \(C=a^2+8a+16=(a+4)^2\). Therefore, the discriminant is \(D=B^2-4AC=4(a+4)^2-4(a+4)^2=0\). Hence, for every real value of \(a\), the roots are real and equal; in fact, the equation is \((x-(a+4))^2=0\), so the repeated root is \(x=a+4\). Exam tip: \(D=0\) indicates equal real roots.

Here (D=4(a-5)^2-4(a^2-10a+21)=16>0). Therefore the roots are real and distinct for every real (a).

A quadratic equation has no real roots when its discriminant satisfies \(D<0\). Here, \(D=[2(k+1)]^2-4(k^2+6k+9)=-16(k+2)\). Therefore, \(-16(k+2)<0\), which gives \(k>-2\). At \(k=-2\), \(D=0\), giving two equal real roots, while \(k<-2\) gives two distinct real roots. Exam tip: simplify the discriminant first and then impose the required sign condition.

Here (D=4(p+1)^2-4(p+4)(p-2)=36>0). Since (p\neq-4), the equation remains quadratic and roots are distinct.

Here, \(a=3\), \(b=-5\), and \(c=2\). The discriminant is \(\Delta=b^2-4ac=(-5)^2-4(3)(2)=25-24=1\). Since \(\Delta>0\), the equation has two distinct real roots. Two equal real roots occur only when \(\Delta=0\), so option B is incorrect. Exam tip: calculate the discriminant first to determine the nature of the roots.

Here (D=6^2-4\cdot1\cdot9=0), so the roots are equal real roots. In exams, (D=0) directly indicates equal roots.

For equal roots, (D=0), so (4^2-4\cdot k\cdot1=0) gives (k=4). In exams, use (D=0) when equal roots are given.

For equal roots, (D=0), so ((-2k)^2-4\cdot1\cdot9=0) gives (k=\pm3). In exams, when (k^2) appears, check both signs.

For two distinct real roots, the discriminant (D=b^2-4ac) is positive. In exams, first calculate (D) to decide the nature.

The direct answer is B: the roots are two equal real roots. For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\), and the roots are \(\frac{-b\pm\sqrt D}{2a}\). If \(D=0\), then \(\sqrt D=0\), so both signs give the same value: \(\frac{-b}{2a}\). This value is real because the coefficients are real and \(a\ne0\). Therefore the equation has a repeated, or equal, real root. Option A is wrong because distinct roots require \(D>0\). Option B is correct. Option C is wrong because no real roots occur when \(D<0\). Option D is too specific and is not guaranteed: the repeated root may be positive, negative, or zero, depending on the coefficients. For example, \(x^2-4x+4=0\) has the equal real root 2. Exam cue: positive discriminant means two distinct real roots, zero means equal real roots, and negative means no real roots.

For equal roots, (D=0), so (m^2-64=0) and (m=\pm8). In exams, remember (m^2=a^2) gives (m=\pm a).

For two distinct real roots, (D>0), so (25-4k>0) gives (k<\frac{25}{4}). In exams, keep the inequality sign correct while solving.

For no real roots, (D<0), so (k^2-64<0), that is (k^2<64). In exams, connect (D<0) with no real roots.

Here (D=2^2-4\cdot7\cdot(-3)=88), so (D>0). In exams, when (c) is negative, (D) can often become positive.

Here (D=12^2-4\cdot9\cdot4=0), so the roots are real and equal. In exams, (9x^2+12x+4) can also be recognized as ((3x+2)^2).

Here (D=(-7)^2-4\cdot1\cdot10=9), which is a positive perfect square. In exams, positive perfect-square (D) gives rational and distinct roots.

Here (D=(-4)^2-4\cdot1\cdot1=12), which is positive but not a perfect square. In exams, such (D) gives irrational distinct real roots.

(D=18) is positive but not a perfect square, so the roots are real, distinct, and irrational. In exams, also check whether (D) is a perfect square.

For equal roots, (D=0), so (q^2-144=0) and (q=\pm12). In exams, both signs of (b) may be possible.

Here (D=(-11)^2-4\cdot6\cdot3=49), a positive perfect square. In exams, roots are rational in such cases.

Here (D=(-4)^2-4\cdot3\cdot2=-8), so there are no real roots. In exams, negative (D) means no intersection on the real number line.

For a quadratic equation to have equal roots, its discriminant must be zero. Here, \(a=1\), \(b=k+2\), and \(c=9\), so \(D=(k+2)^2-4(1)(9)=0\). Therefore, \((k+2)^2=36\), giving \(k+2=\pm6\) and hence \(k=4\) or \(k=-8\). Option B results from reversing the signs incorrectly. Exam tip: For equal roots, begin by setting the discriminant \(D\) equal to zero.

For equal roots, (D=0), so (100-4t=0) gives (t=25). In exams, (b^2=4ac) is a quick sign of equal roots.

Here (D=(-7)^2-4\cdot2\cdot6=1), so the roots are two distinct real roots. In exams, (D=1) is also a positive perfect square.

The direct answer is A: k=0 or k=1. For 4x^2+4kx+k=0, identify a=4, b=4k, and c=k. Equal roots require D=b^2-4ac=0. Substitute: D=(4k)^2-4(4)(k)=16k^2-16k=16k(k-1). Set it equal to zero: 16k(k-1)=0. Since 16 is nonzero, k=0 or k-1=0, so k=0 or k=1. Also a=4 is never zero, so the equation remains quadratic for both values. Option A is correct. Option B omits k=0 and adds k=2. Option C adds k=4, which does not satisfy the condition. Option D includes k=-1, also not a solution. Use D=0 for equal roots.