Nature of Roots

Here (D=(-2\ell)^2-4(\ell^2-9)=36>0). So for every real (\ell), roots are real and distinct.

Here, \(a=4\), \(b=-4w\), and \(c=w^2\). Therefore, the discriminant is \(D=b^2-4ac=(-4w)^2-4(4)(w^2)=0\). Hence, for every real \(w\), the roots are real and equal; in fact, the equation is \((2x-w)^2=0\), giving the repeated root \(x=\frac{w}{2}\). Even when \(w=0\), the coefficient of \(x^2\) remains 4, so the equation is still quadratic. Exam tip: When \(D=0\), the roots are real and equal.

Here, \(a=2\), \(b=-4v\), and \(c=2v^2+5\). Therefore, the discriminant is \(D=b^2-4ac=(-4v)^2-4(2)(2v^2+5)=16v^2-16v^2-40=-40\). Since \(D<0\) for every real value of \(v\), the equation has no real roots. Hence, option A is correct; option D is incorrect because the nature of the roots does not change with \(v\). Exam tip: a quadratic equation has no real roots when its discriminant is negative.

For a quadratic equation to have real roots, its discriminant must satisfy \(D\ge0\). Here, \(a=1\), \(b=2g\), and \(c=g^2-6g+11\). Thus, \(D=(2g)^2-4(g^2-6g+11)=24g-44\). Therefore, \(24g-44\ge0\), giving \(g\ge\frac{11}{6}\). At \(g=\frac{11}{6}\), the roots are equal and real, so option B is incorrect because it excludes this boundary value. Exam tip: For questions about real roots, begin by applying \(D\ge0\).

For equal roots, the discriminant \(D=b^2-4ac\) must be zero. Here, \(a=1\), \(b=2g\), and \(c=g^2-6g+11\). Therefore, \(D=(2g)^2-4(g^2-6g+11)=24g-44\). Setting \(D=0\) gives \(24g-44=0\), so \(g=\frac{11}{6}\). Option B results from incorrectly reversing the numerator and denominator. In an exam, first apply the condition \(D=0\) for equal roots.

A quadratic equation has real and distinct roots only when its discriminant satisfies \(D>0\). Here, \(a=1\), \(b=-2h\), and \(c=h^2+8h\), so \(D=b^2-4ac=4h^2-4(h^2+8h)=-32h\). Therefore, \(-32h>0\), which gives \(h<0\). Note that \(h=0\) makes \(D=0\), producing equal roots, while \(h>0\) gives non-real roots. Exam tip: For real and distinct roots, always impose \(D>0\).

For equal roots, the discriminant must be zero: \(D=b^2-4ac=0\). Here, \(a=1\), \(b=-2h\), and \(c=h^2+8h\). Thus, \(D=(-2h)^2-4(1)(h^2+8h)=-32h\). Setting \(-32h=0\) gives \(h=0\). Indeed, for this value the equation becomes \(x^2=0\), whose two roots are both \(0\). Exam tip: For equal roots of a quadratic equation, immediately use the condition \(D=0\).

For a quadratic equation \(Ax^2+Bx+C=0\) to have real roots, its discriminant must satisfy \(D=B^2-4AC\ge0\). Here, \(A=3\), \(B=-2(2a+1)\), and \(C=a^2+a+1\), so \(D=4(2a+1)^2-12(a^2+a+1)=4(a^2+a-2)\). Therefore, the required condition is \(a^2+a-2\ge0\). Exam tip: apply the discriminant condition first and solve the resulting inequality only if the question asks for the range of \(a\).

A quadratic equation has real and distinct roots when its discriminant satisfies \(D>0\). Here, \(A=3\), \(B=-2(2a+1)\), and \(C=a^2+a+1\). Thus, \(D=B^2-4AC=4(a^2+a-2)=4(a+2)(a-1)\). Therefore, \((a+2)(a-1)>0\), which gives \(a<-2\) or \(a>1\). At \(a=-2\) or \(a=1\), \(D=0\), so the roots are equal rather than distinct. Exam tip: \(D>0\), \(D=0\), and \(D<0\) indicate real distinct, real equal, and non-real roots, respectively.

A quadratic equation has no real roots when its discriminant satisfies \(D<0\). Here, \(D=[-2(2a+1)]^2-4(3)(a^2+a+1)=4(a^2+a-2)=4(a+2)(a-1)\). Therefore, \((a+2)(a-1)<0\), which gives \(-2<a<1\). At the endpoints \(a=-2\) and \(a=1\), \(D=0\), so they are not included. Exam tip: \(D<0\) indicates no real roots, whereas \(D=0\) indicates equal real roots.

Since D = 18 > 0, the roots are real and distinct. Also, 18 is not a perfect square, and √18 = 3√2 is irrational. Therefore, for a quadratic equation with rational coefficients, the roots are irrational. The student’s statement that the roots are rational is incorrect. Exam tip: D > 0 indicates real and distinct roots, while whether D is a perfect square determines their rationality.

For a quadratic equation, the discriminant is D = b² − 4ac. When D < 0, the equation has no real roots; here, D = −4. Option C is incorrect because equal real roots require D = 0, while two distinct real roots require D > 0. Exam tip: check the sign of the discriminant—negative, zero and positive indicate no real roots, equal roots and distinct real roots, respectively.

For every real (a), ((a-1)^2\geq 0). Hence, (D=(a-1)^2+5\geq 5>0). A quadratic equation with (D>0) has two real and distinct roots. Equal roots require (D=0), which is impossible here. Exam tip: use the discriminant directly—(D>0) means distinct real roots, (D=0) means equal real roots, and (D<0) means non-real roots.

The roots of a quadratic equation are real and equal only when its discriminant is \(D=0\). Thus, \(-(b+3)^2=0\), which gives \((b+3)^2=0\) and hence \(b=-3\). Therefore, option A is correct. Exam tip: the negative of a real square can be zero only when the quantity being squared is zero.

Since \(b\neq -3\), we have \(b+3\neq 0\), so \((b+3)^2>0\). Therefore, \(D=-(b+3)^2<0\). For a quadratic equation, a negative discriminant means that the roots are not real; they are complex conjugates. Hence, option A is correct. Exam tip: Always remember that \(D<0\) means no real roots.

For the equation x² − (2r+5)x + (r²+5r+4) = 0, the coefficients are a = 1, b = −(2r+5) and c = r²+5r+4. Its discriminant is D = [−(2r+5)]² − 4(r²+5r+4) = (2r+5)² − 4r² − 20r − 16. Expanding the square gives 4r²+20r+25, so D = 9. Since D is always positive, the roots are always real and distinct. Also, D = 9 is a perfect square and the coefficients are rational whenever r is rational; algebraically the roots simplify directly to r+4 and r+1. Thus the intended school-level conclusion is that the roots are real, rational and distinct, so option A is correct. They are not equal or non-real.

For a quadratic equation, the discriminant is \(D=b^2-4ac\). Here, \(a=1\), \(b=-(2r+5)\), and \(c=r^2+5r+7\). Thus, \(D=(2r+5)^2-4(r^2+5r+7)=-3\). Since \(D<0\) for every real value of \(r\), the equation has no real roots. Therefore, options B and C are impossible, while option D is incorrect because the result is not restricted to \(r=0\). Exam tip: whenever \(D<0\), conclude immediately that the quadratic has no real roots.

For option A, the discriminant is \(D=(2m)^2-4(m^2+1)=-4\), which is negative for every real \(m\). Hence its roots are non-real. In option B, \(D=4\). Exam tip: determine the sign of the discriminant first.

Here, \(a=1\), \(b=-2(k-3)\), and \(c=k^2-6k+8\). Therefore, the discriminant is \(D=b^2-4ac=4(k-3)^2-4(k^2-6k+8)=4\). Since \(D=4>0\) for every real value of \(k\), the roots are always real and distinct. Exam tip: \(D>0\) indicates two real and unequal roots, \(D=0\) indicates equal roots, and \(D<0\) indicates non-real roots.

Here, a=1, b=-2(k-3), and c=k^2-6k+9. Therefore, the discriminant is \(D=b^2-4ac=4(k-3)^2-4(k^2-6k+9)=0\), since \(k^2-6k+9=(k-3)^2\). Thus, for every real value of k, the roots are real and equal; in fact, the equation is \((x-(k-3))^2=0\), giving the repeated root \(x=k-3\). Exam tip: \(D=0\) indicates equal real roots.