Nature of Roots

Here, \(a=1\), \(b=-(r+5)\), and \(c=5r\). Therefore, the discriminant is \(D=b^2-4ac=(r+5)^2-20r=(r-5)^2\). Two real and distinct roots require \(D>0\). Hence, \((r-5)^2>0\), which gives \(r\ne5\). In option B, \(r=5\) makes \(D=0\), so the roots are equal. Exam tip: For two distinct real roots of a quadratic equation, always check that \(D>0\).

Here, \(a=1\), \(b=-(s+4)\), and \(c=4s\). Equal roots require the discriminant \(D=b^2-4ac\) to be zero. Thus, \(D=(s+4)^2-16s=s^2-8s+16=(s-4)^2\). Therefore, \((s-4)^2=0\), giving \(s=4\). The distractor \(s=0\) is incorrect because it gives \(D=16\), not zero. Exam tip: for equal roots, immediately apply the condition \(D=0\).

For a quadratic equation \\(ax^2+bx+c=0\\), equal roots occur when the discriminant \\(D=b^2-4ac\\) is zero. Here, \\(a=q+2\\), \\(b=-2(q-1)\\), and \\(c=q\\). Thus, \\(D=4(q-1)^2-4q(q+2)=4(1-4q)\\). Setting \\(D=0\\) gives \\(1-4q=0\\), so \\(q=\\frac14\\). At \\(q=1\\), the discriminant is not zero, while \\(q=-2\\) removes the quadratic term and is also excluded by the condition. Exam tip: For equal roots, begin by setting the discriminant equal to zero.

Here, \(a=1\), \(b=2(t+1)\), and \(c=3t+7\). Therefore, the discriminant is \(D=b^2-4ac=4(t+1)^2-4(3t+7)=4(t-3)(t+2)\). An equation has no real roots when \(D<0\), so \((t-3)(t+2)<0\), giving \(-2<t<3\). Option B is incorrect because it does not come from the zeros \(-2\) and \(3\); at the endpoints \(t=-2,3\), \(D=0\) and the equation has equal real roots. Exam tip: For a quadratic equation, use \(D<0\) to identify the condition for no real roots.

For the equation \\(x^2-2(a+b)x+2ab=0\\), the coefficients are 1, \\(-2(a+b)\\) and 2ab. Its discriminant is \\(D=[-2(a+b)]^2-8ab=4(a+b)^2-8ab=4(a^2+b^2)\\). Since squares are non-negative, \\(a^2+b^2\geq0\\), so D is always non-negative.

Therefore the equation always has real roots, including the possibility of equal roots when both a and b are zero. Option A expresses this correctly. The condition ab>0 is not necessary; a and b may have opposite signs or one may be zero. Also, equal roots require D=0, namely a=b=0, not merely a+b=0.

The discriminant of the quadratic equation is \(D=[-2(a-b)]^2-4(a+b)^2\). Therefore, \(D=4[(a-b)^2-(a+b)^2]=-16ab\). For real roots, \(D\geq0\), so \(-16ab\geq0\), which gives \(ab\leq0\). Option B reverses the inequality. The condition \(a+b=0\) may satisfy the requirement, but it is not the complete necessary condition. Exam tip: begin with \(D\geq0\) whenever the nature of roots is asked.

Since \(a>0\) and \(c<0\), we have \(ac<0\). Therefore, \(-4ac>0\), and for any real \(b\), the discriminant \(D=b^2-4ac=b^2+(-4ac)>0\). Hence, the equation has two real and distinct roots. Equal roots would require \(D=0\), which is impossible here. Exam tip: When \(ac<0\), the roots are always real and distinct.

Here, a=7, b=-2\(\sqrt{21}\), and c=3. Therefore, the discriminant is \(D=b^2-4ac=(-2\sqrt{21})^2-4(7)(3)=84-84=0\). Hence, the roots are real and equal. In fact, the repeated root is \(x=\frac{\sqrt{21}}{7}\). Option D incorrectly treats 84, which is only the value of \(b^2\), as the discriminant. Exam tip: \(D=0\) always indicates two equal real roots.

A quadratic equation has two real and distinct roots only when its discriminant satisfies \(D>0\). Here, \(a=1\), \(b=-2\mu\), and \(c=2\mu\), so \(D=b^2-4ac=4\mu^2-8\mu=4\mu(\mu-2)\). Therefore, \(4\mu(\mu-2)>0\), which gives \(\mu<0\) or \(\mu>2\). At \(\mu=0\) or \(\mu=2\), \(D=0\), so the roots are real but equal, not distinct. Exam tip: For two real and unequal roots, always apply the condition \(D>0\).

Here, \(a=1\), \(b=-2\mu\), and \(c=2\mu\). A quadratic equation has equal roots when its discriminant \(D=b^2-4ac\) is zero. Thus, \(D=4\mu^2-8\mu=4\mu(\mu-2)=0\), giving \(\mu=0\) or \(\mu=2\). Hence, option A is correct. Choosing only \(\mu=2\) or only \(\mu=0\) omits one valid value. Exam tip: For equal roots, set the discriminant equal to zero.

Here (D=4\alpha^2-4(\alpha+2)(\alpha-1)=8-4\alpha). For real roots (\alpha\leq2), and for a quadratic (\alpha\neq-2).

A quadratic equation has no real roots when its discriminant satisfies \(D<0\). Here, \(a=1\), \(b=-2(3t+1)\), and \(c=5t^2+2t+4\). Thus, \(D=4(3t+1)^2-4(5t^2+2t+4)=4(2t-1)(2t+3)\). Therefore, \((2t-1)(2t+3)<0\), which gives \(-\frac{3}{2}<t<\frac{1}{2}\). At the endpoints, \(D=0\), so the equation has equal real roots rather than no real roots. Exam tip: for a positive leading coefficient, a factored quadratic is negative between its two distinct roots.

For a quadratic equation \(ax^2+bx+c=0\) to have equal roots, its discriminant must be zero: \(D=b^2-4ac=0\). Here, \(a=1\), \(b=2(m-4)\), and \(c=m^2-7m+14\). Therefore, \(D=4(m-4)^2-4(m^2-7m+14)=4(2-m)\). Setting \(D=0\) gives \(4(2-m)=0\), so \(m=2\). Exam tip: whenever equal roots are mentioned, immediately use the condition \(D=0\).

Here, \(a=1\), \(b=2(m-4)\), and \(c=m^2-7m+14\). Therefore, the discriminant is \(D=b^2-4ac=4(m-4)^2-4(m^2-7m+14)=4(2-m)\). For real roots, \(D\geq0\), so \(4(2-m)\geq0\), which gives \(m\leq2\). Remember that \(D=0\) gives two equal real roots, so the boundary value \(m=2\) is included.

Here (D=4(3k-1)^2-8(k+1)^2). For equal roots, solve (D=0) directly and avoid mental shortcuts.

For \(ax^2+bx+c=0\), the product of the roots is \(c/a\). In option A, \(c/a=-6/1=-6\), which is negative, so the roots have opposite signs. In option C, the product is \(6\). Exam tip: check \(c/a<0\) for roots of opposite signs.

Here, \(a=1\), \(b=-2p\), and \(c=p^2+16\). Therefore, the discriminant is \(D=b^2-4ac=(-2p)^2-4(1)(p^2+16)=4p^2-4p^2-64=-64\). Since \(D<0\) for every real value of \(p\), the equation has no real roots. Option B would require \(D=0\), which is not the case here. Exam tip: a negative discriminant means that a quadratic equation has no real roots.

Here, \(A=1\), \(B=-2(a-b)\), and \(C=(a+b)^2\). Therefore, \(D=B^2-4AC=4(a-b)^2-4(a+b)^2=-16ab\). Since \(ab<0\), we have \(-16ab>0\), so the discriminant is positive and the roots are real and distinct. Hence, both the assertion and the reason are correct, and the reason explains the assertion. Exam tip: For a quadratic equation, \(D>0\) indicates two real and distinct roots.

Here, \(a=2\), \(b=-4\), and \(c=7\). Thus, \(D=b^2-4ac=(-4)^2-4(2)(7)=16-56=-40\). Since \(D<0\), the equation has no real roots; therefore, the parabola \(y=2x^2-4x+7\) does not intersect the \(x\)-axis. Hence, both the assertion and the reason are correct, and the reason explains the assertion. Exam tip: If \(D<0\), the graph has no real point of intersection with the \(x\)-axis.

Equal roots require the discriminant to be zero. Here, \(D=(m-4)^2-4m=m^2-12m+16\). Setting \(D=0\) gives \(m=6\pm2\sqrt5\). At \(m=4\), \(D=-16\), so there are no real roots. Exam tip: never decide the nature of roots merely from the coefficient of \(x\).

A quadratic equation has equal roots when its discriminant is \(D=0\). Thus, \((w+2)^2=0\), which gives \(w+2=0\) and hence \(w=-2\). The distractor \(w=2\) gives \(D=16\), leading to two distinct real roots, not equal roots. Exam tip: remember that \(D=0\) indicates equal roots, \(D>0\) indicates two distinct real roots, and \(D<0\) indicates no real roots.

A quadratic equation has no real roots when its discriminant satisfies \(D<0\). Thus, we solve \((z-4)(z+6)<0\). The product is negative between its zeros, \(-6\) and \(4\), so the correct interval is \(-6<z<4\). In option B, the product is positive, giving two real roots instead. Exam tip: for a product of two linear factors with positive leading coefficient, the sign is negative between the two zeros.

A quadratic equation has two real and distinct roots only when its discriminant satisfies \(D>0\). Therefore, \(12n-36>0\), which gives \(12n>36\) and hence \(n>3\). At \(n=3\), \(D=0\), so the roots are real but equal. Exam tip: remember \(D>0\) for two real and distinct roots.

A quadratic equation has equal roots when its discriminant is zero. Thus, \(36-9h^2=0\), giving \(9h^2=36\) and hence \(h^2=4\). Therefore, \(h=\pm2\), so option A is correct. Option B results from taking the square root of 36 without properly dividing by 9. Exam tip: For equal roots, always set \(b^2-4ac=0\).

Here, \(a=1\), \(b=2(2-\sqrt{5})\), and \(c=9\). Therefore, the discriminant is \(D=b^2-4ac=4(2-\sqrt{5})^2-36=4(9-4\sqrt{5})-36=-16\sqrt{5}<0\). Since \(D<0\), the equation has no real roots. Option B is incorrect because two equal real roots require \(D=0\). Exam tip: for a quadratic equation, \(D<0\) indicates that the roots are non-real.