Nature of Roots

For real and equal roots, the discriminant must be zero, and the coefficient of x² must remain non-zero. Here a = λ+1, b = −2(λ−2) and c = λ+1. Thus D = b² − 4ac = 4(λ−2)² − 4(λ+1)². Using the difference of squares, D = 4[(λ−2)² − (λ+1)²] = 4[(−3)(2λ−1)] = 12(1−2λ). Setting D = 0 gives 1−2λ = 0, so λ = 1/2. At this value, a = 3/2, which is non-zero, so the equation is genuinely quadratic and has equal real roots. Therefore option A is correct. λ = −1 would remove the quadratic term, while the other values do not make the discriminant zero.

Here (D=4(\mu+3)^2-8(\mu^2+6\mu+5)=-4(\mu^2+6\mu+1)). It is not always positive, so all (\mu) is not correct.

For a quadratic equation, the discriminant is \(D=b^2-4ac\). Here, \(a_1=1\), \(b=-2(a+4)\), and \(c=a^2+8a+20\). Thus, \(D=4(a+4)^2-4(a^2+8a+20)=-16\), which is negative for every real value of \(a\). Therefore, the equation has no real roots, so their number is \(0\). Exam tip: \(D<0\) means no real roots, whereas \(D=0\) means one repeated real root.

Here \(A=1\), \(B=-2(a+4)\), and \(C=a^2+8a+15\). Therefore, the discriminant is \(D=B^2-4AC=4(a+4)^2-4(a^2+8a+15)=4>0\). Hence, the roots are always real and distinct. In fact, the roots are \(a+3\) and \(a+5\); thus, they are rational when \(a\) is rational. Option B is incorrect because \(D\neq0\), and option D is incorrect because rational values of \(a\) give rational roots. Exam tip: \(D>0\) indicates real and distinct roots, \(D=0\) indicates equal roots, and \(D<0\) indicates non-real roots.

Here, \(A=1\), \(B=-2(a+4)\), 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, \(a=2, b=-11, c=15\), so the discriminant is \(\Delta=b^2-4ac=(-11)^2-4(2)(15)=1\). Since \(\Delta\) is positive and a perfect square, the roots are real, rational, and distinct. In fact, \(2x^2-11x+15=(2x-5)(x-3)\), giving the roots \(x=\frac{5}{2}\) and \(x=3\). Exam tip: when \(\Delta>0\) and is a perfect square, the roots are rational and distinct.

Here, \(a=1, b=10, c=29\). Therefore, the discriminant is \(D=b^2-4ac=10^2-4(1)(29)=100-116=-16\). Since \(D<0\), the equation has no real roots. Option B is incorrect because equal real roots require \(D=0\). Exam tip: To determine the nature of roots, first calculate the discriminant and check its sign.

A quadratic equation \(ax^2+bx+c=0\) has real roots when its discriminant \(D=b^2-4ac\) is non-negative. Here, \(a=1, b=-2k, c=9\), so \(D=(-2k)^2-4(1)(9)=4k^2-36\). Thus, \(4k^2-36\geq0\), which gives \(k^2\geq9\), or \(k\leq-3\) or \(k\geq3\). Option B makes the discriminant negative, while option D is incomplete because \(k=-3\) also gives real equal roots. Exam tip: For real roots, always apply the condition \(D\geq0\).

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.

For a quadratic equation ax²+bx+c=0, the discriminant is D=b²−4ac. Its sign determines the nature of the roots. When D>0, the square root of D is real and nonzero, so the quadratic formula gives two real and distinct roots: x=(-b+√D)/(2a) and x=(-b−√D)/(2a). If D is also a perfect square, these roots are rational, assuming rational coefficients in the usual school setting. However, because D is not a perfect square, √D is irrational, making both roots irrational. Hence option A is correct. Options B, C, and D correspond respectively to a square positive discriminant, zero discriminant, and negative discriminant.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=1,b=-1,c=-1\), so \(D=(-1)^2-4(1)(-1)=5\). Since \(D>0\), the roots are real and distinct; since 5 is not a perfect square, the roots are irrational. Therefore, option A is correct. Exam tip: If \(D>0\) and \(D\) is not a perfect square, the roots are real, distinct, and irrational.

For the given equation, \(a=2\), \(b=-1\), and \(c=2\). Thus, the discriminant is \(D=b^2-4ac=(-1)^2-4(2)(2)=1-16=-15\). Since \(D<0\), the quadratic equation has no real roots. Exam tip: calculate the discriminant first and identify the nature of the roots from its sign.

The equation can be factorised as \(9x^2-6x+1=(3x-1)^2\). Thus, \((3x-1)^2=0\), giving \(3x-1=0\) and the equal root \(x=\frac{1}{3}\). Exam tip: For equal roots, the discriminant \(D=b^2-4ac\) is zero.

Here, \(a=4, b=7, c=-2\). Therefore, the discriminant is \(D=b^2-4ac=7^2-4(4)(-2)=81\). Since \(D\) is positive and a perfect square, the roots are real, rational, and distinct. In fact, the roots are \(\frac{1}{4}\) and \(-2\). Option B is incorrect because a positive discriminant gives irrational roots only when it is not a perfect square. Exam tip: If \(D>0\) and is a perfect square, the roots are real, rational, and distinct.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=1, b=11, c=31\), so \(D=11^2-4(1)(31)=121-124=-3\). Since \(D<0\), the equation has no real roots. Exam tip: two real and equal roots occur only when \(D=0\), not when the discriminant is negative.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=3\), \(b=-2\), and \(c=-2\), so \(D=(-2)^2-4(3)(-2)=28\). Since \(D>0\), the roots are real and distinct; since 28 is not a perfect square, the roots are irrational. Option B is incorrect because the discriminant is 28, not 4. Exam tip: If \(D>0\) and is not a perfect square, the roots are real, distinct, and irrational.

Here, \\(a=16, b=24, c=9\\). The discriminant is \\(D=b^2-4ac=24^2-4(16)(9)=0\\), so the roots are real and equal. Also, \\(16x^2+24x+9=(4x+3)^2\\), which gives \\(4x+3=0\\) and hence the equal root \\(x=-\frac{3}{4}\\). Therefore, option B has the wrong sign, while option C gives an incorrect discriminant. Exam tip: If a quadratic is a perfect square or its discriminant is zero, its roots are equal.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=5, b=1, c=1\), so \(D=1^2-4(5)(1)=-19\). Since \(D<0\), the equation has no real roots; therefore, option A is correct. Exam tip: \(D<0\) means no real roots, \(D=0\) means equal real roots, and \(D>0\) means two distinct real roots.