Nature of Roots

For a quadratic equation, the discriminant is \\(D=b^2-4ac\\). Here, \\(a=8, b=-4\sqrt{10}, c=5\\), so \\(D=(-4\sqrt{10})^2-4(8)(5)=160-160=0\\). Therefore, the roots are real and equal. In fact, the repeated root is \\(x=\frac{\sqrt{10}}{4}\\), which is irrational; however, option D is still incorrect because it says the roots are distinct. Exam tip: \\(D=0\\) always indicates two equal real roots.

A quadratic equation has two real and distinct roots only when its discriminant is positive. Here, \(a=1\), \(b=-2\theta\), and \(c=3\theta\), so \(D=b^2-4ac=4\theta^2-12\theta=4\theta(\theta-3)\). Therefore, \(4\theta(\theta-3)>0\), which gives \(\theta<0\) or \(\theta>3\). At \(\theta=0\) and \(\theta=3\), the roots are equal, while for \(0<\theta<3\), the roots are not real. Exam tip: For two real and distinct roots, always impose \(D>0\).

For equal roots, the discriminant \(D=b^2-4ac\) must be zero. Here, \(a=1\), \(b=-2\theta\), and \(c=3\theta\), so \(D=(-2\theta)^2-4(1)(3\theta)=4\theta(\theta-3)\). Thus, \(4\theta(\theta-3)=0\) gives \(\theta=0\) or \(\theta=3\). Option B is incomplete because it omits \(\theta=0\). Exam tip: For equal roots, set the discriminant directly to zero and check every resulting parameter value.

Here (D=4\alpha^2-4(\alpha+3)(\alpha-2)=24-4\alpha). For real roots (\alpha\leq3), and for a quadratic (\alpha\neq-3).

For a quadratic equation, real roots do not exist when the discriminant \(D<0\). Here, \(D=[-2(4t+1)]^2-4(7t^2+2t+5)=4(9t^2+6t-4)\). Thus, \(9t^2+6t-4<0\). Its zeros are \(\frac{-1-\sqrt5}{3}\) and \(\frac{-1+\sqrt5}{3}\), and since the quadratic has a positive leading coefficient, the expression is negative between these zeros. Hence option A is correct. At the endpoints, \(D=0\), giving equal real roots, so the endpoints must be excluded. Exam tip: for no real roots, always use the strict condition \(D<0\), not \(D\leq0\).

For a quadratic equation \(ax^2+bx+c=0\) to have equal roots, its discriminant must be zero: \(D=b^2-4ac=0\). Here, \(a=1\), \(b=2(m-5)\), and \(c=m^2-9m+24\). Thus, \(D=4(m-5)^2-4(m^2-9m+24)=4(1-m)\). Setting \(D=0\) gives \(4(1-m)=0\), so \(m=1\). Exam tip: For equal-root questions, begin with \(D=0\) and simplify the discriminant carefully.

Here, a = 1, b = 2(m − 5), and c = m² − 9m + 24. Therefore, the discriminant is D = b² − 4ac = 4(m − 5)² − 4(m² − 9m + 24) = 4(1 − m). For the equation to have no real roots, D must be less than zero. Thus, 4(1 − m) < 0 gives m > 1. When m = 1, D = 0, so the equation has two equal real roots; hence option C is incorrect. Exam tip: For a quadratic equation, remember that no real roots occur when D < 0.

Here \(a=1, b=2p\), and \(c=p^2+1\). Thus \(\Delta=b^2-4ac=4p^2-4(p^2+1)=-4<0\), so there are no real roots. In exams, check the sign of the discriminant first.

Here, \(a=1\), \(b=-2p\), and \(c=p^2+36\). Therefore, the discriminant is \(D=b^2-4ac=(-2p)^2-4(p^2+36)=-144\), which remains negative for every real value of \(p\). Hence, the equation has no real roots. Equal real roots would require \(D=0\), whereas here \(D<0\). Exam tip: For a quadratic equation, \(D<0\) immediately indicates that there are no real roots.

Here, \(A=1\), \(B=-2(a-3b)\), and \(C=(a+3b)^2\). Thus, \(D=B^2-4AC=4(a-3b)^2-4(a+3b)^2=-48ab\). Since \(ab<0\), we have \(-48ab>0\), so the roots are real and distinct. The roots would be real and equal only if \(D=0\). Exam tip: First determine the sign of the discriminant—\(D>0\) indicates real and distinct roots.

Here, \(a=3\), \(b=-6\), and \(c=11\). Thus, the discriminant is \(D=b^2-4ac=(-6)^2-4(3)(11)=36-132=-96\). Since \(D<0\), the equation has no real roots; therefore, the parabola neither intersects nor touches the \(x\)-axis. Hence, both the assertion and the reason are correct, and the reason correctly explains the assertion. Exam tip: \(D<0\) indicates no real intersection between the parabola and the \(x\)-axis.

Here, \(a=1\), \(b=-2(k-3)\), and \(c=k^2\). Thus, \(D=b^2-4ac=[-2(k-3)]^2-4(1)(k^2)=4(k-3)^2-4k^2=36-24k\). Option B has the wrong sign for the linear term, while options C and D result from errors in expanding the square or handling its coefficients. Exam tip: identify the complete coefficient \(b\) before substituting into \(b^2-4ac\).

Since \\(y\neq -5\\), we have \\(y+5\neq 0\\), so \\(D_1=(y+5)^2>0\\). A quadratic equation has two distinct real roots when its discriminant satisfies \\(D>0\\). In contrast, \\(D_2=-(y+5)^2<0\\), which gives no real roots. Exam tip: Check the sign of the discriminant first to determine the nature of the roots.

A quadratic equation has equal roots only when its discriminant is \(D=0\). Therefore, \((m-8)^2=0\), which gives \(m-8=0\) and hence \(m=8\). Option D is incorrect because \((m-8)^2\) is not zero for every real value of \(m\). Exam tip: For equal-root questions, begin by setting the discriminant equal to zero.

A quadratic equation has no real roots when its discriminant satisfies D < 0. Therefore, (z − 9)(z + 1) < 0. This product is negative between its zeros, −1 and 9, so −1 < z < 9. In option B, D > 0, which gives two distinct real roots; at z = −1 or z = 9, D = 0, which gives equal real roots. Exam tip: For a product of two linear factors, test a value such as z = 0 to confirm the sign in the interval between the zeros.

A quadratic equation has two real and distinct roots only when its discriminant satisfies \(D>0\). Thus, \(20n-80>0\), so \(20n>80\) and hence \(n>4\). At \(n=4\), \(D=0\), which gives two equal real roots, not distinct roots. Exam tip: check the sign of the discriminant first—\(D>0\) indicates real and distinct roots.

The discriminant is \\(D=b^2-4ac=4(3-\sqrt{10})^2-64=12-24\sqrt{10}=12(1-2\sqrt{10})<0\\). Therefore, the equation has no real roots. Option B would require \\(D=0\\), but the discriminant here is negative. Exam tip: For a quadratic equation, \\(D<0\\) means that the roots are not real.

Here, \(a=1\), \(b=-2(3+\sqrt{5})\), and \(c=14+6\sqrt{5}\). Therefore, the discriminant is \(D=b^2-4ac=4(3+\sqrt{5})^2-4(14+6\sqrt{5})=0\), since \((3+\sqrt{5})^2=14+6\sqrt{5}\). Hence, the roots are real and equal. In fact, the repeated root is \(3+\sqrt{5}\), which is irrational; therefore, option D is incorrect because it says the roots are distinct. Exam tip: For a quadratic equation, \(D=0\) always indicates two real and equal roots.

The discriminant is \(D=(-2r)^2-4(r^2-81)=4r^2-4r^2+324=324=18^2\). Since \(D>0\), the equation has two real and distinct roots for every real value of \(r\). In fact, the roots are \(r+9\) and \(r-9\). They cannot be called always rational because \(r\) itself may be irrational, so option D is not correct. Exam tip: \(D>0\), \(D=0\), and \(D<0\) indicate distinct real, equal real, and non-real roots, respectively.

For a quadratic equation, the discriminant is \(D=b^2-4ac\). Here, \(a=1\), \(b=-2r\), and \(c=r^2+49\), so \(D=(-2r)^2-4(1)(r^2+49)=-196<0\). Therefore, for every real value of \(r\), the equation has no real roots. Option B would require \(D=0\), which is not possible here. Exam tip: a quadratic equation with real coefficients has no real roots whenever its discriminant is negative.

On the \(x\)-axis, \(y=0\). Therefore, the intersection equation is \(x^2-2kx+k^2+9=0\), or \((x-k)^2+9=0\). Since \((x-k)^2\geq 0\), the left side is always at least 9 and can never be zero. Equivalently, the discriminant is \(D=(-2k)^2-4(k^2+9)=-36<0\), so there are no real points of intersection. The tempting answers of one or two intersections are incorrect because they would require a zero or positive discriminant. Exam tip: completing the square gives the result immediately.

The equation can be written as \(y=(x-(k+2))^2\). On the \(x\)-axis, \(y=0\), so \((x-(k+2))^2=0\), which has the single real root \(x=k+2\). Therefore, for every real value of \(k\), the parabola only touches the \(x\)-axis; it is not restricted to \(k=-2\). Exam tip: when the discriminant of the corresponding quadratic is \(D=0\), the graph touches the axis at exactly one point.

To find intersections with the \(x\)-axis, set \(y=0\), giving \(x^2-2kx+k^2-16=0\). Its discriminant is \(D=(-2k)^2-4(k^2-16)=64>0\). Therefore, for every real value of \(k\), the equation has two distinct real roots, so the parabola intersects the \(x\)-axis at two distinct points. Exam tip: \(D>0\) indicates two distinct real roots.

By Vieta’s relation, the product of the roots is \(\frac{c}{a}\). A negative product gives opposite signs. Also, \(ac<0\) makes \(b^2-4ac>0\), so both roots are real. Exam tip: use the sign of the product to identify root signs.

By Vieta’s relation, the product of the roots is \(\alpha\beta=\frac{c}{a}\). If \(\frac{c}{a}<0\), one root is positive and the other is negative. For \(\frac{c}{a}>0\), real roots have the same sign. Exam tip: also check \(b^2-4ac>0\) for distinct real roots.