Nature of Roots

Here, \(a=1\), \(b=-2p\), and \(c=p^2-q^2\). Therefore, the discriminant is \(D=b^2-4ac=4p^2-4(p^2-q^2)=4q^2\). Since \(q\neq 0\), we have \(D=4q^2>0\), so the equation has two real and distinct roots. In fact, the roots are \(p+q\) and \(p-q\). Option B is incorrect because equal roots require \(D=0\). Exam tip: For a quadratic equation, \(D>0\) always indicates two real and distinct roots.

Here, \(a=1\), \(b=-2p\), and \(c=p^2+q^2\). Therefore, the discriminant is \(D=b^2-4ac=4p^2-4(p^2+q^2)=-4q^2\). Since \(q\ne0\), \(D<0\), so the equation has no real roots. In fact, its roots are \(p\pm iq\). Exam tip: For a quadratic equation, \(D<0\) immediately implies that the roots are not real.

Here, \(A=1\), \(B=-2(a+b)\), and \(C=(a-b)^2\). Thus, \(D=B^2-4AC=4(a+b)^2-4(a-b)^2=16ab\). Since \(ab>0\), we have \(D>0\), so the roots are real and distinct. Therefore, the reason is correct and directly explains the assertion. Exam tip: Determine the sign of \(D\) first; \(D>0\) indicates two real and distinct roots.

Here, \(a=1\), \(b=-2(k+1)\), and \(c=k^2\). Hence, \(D=b^2-4ac=4(k+1)^2-4k^2=4[(k+1)^2-k^2]=4(2k+1)\). Option B ignores the terms involving \(k\), while option D omits the constant term \(4\). Exam tip: identify \(a,b,c\) carefully before applying the discriminant formula, and simplify only afterward.

For a quadratic equation \(ax^2+bx+c=0\), equal roots occur if and only if the discriminant \(D=b^2-4ac\) is zero. In option A, \(a=7\), \(b=-10\sqrt{7}\), and \(c=25\), so \(D=(-10\sqrt{7})^2-4(7)(25)=700-700=0\). Hence, its two roots are equal. For instance, option B has \(D=(-9\sqrt{7})^2-700=-133\), so it does not have real roots. Exam tip: To identify equal roots, check \(D=0\) directly instead of using the full quadratic formula.

For option A, the discriminant is \(D=b^2-4ac=(-2\sqrt{2})^2-4(1)(-1)=8+4=12\). Since \(D>0\), the roots are real and distinct. Also, \(\sqrt{D}=\sqrt{12}=2\sqrt{3}\) is irrational, giving the roots \(\sqrt{2}+\sqrt{3}\) and \(\sqrt{2}-\sqrt{3}\), both of which are irrational. In option B, the discriminant is zero; option C has a negative discriminant; and option D has the rational roots 2 and 3. Exam tip: real and distinct roots require \(D>0\), while irrational roots require the square root of the discriminant to be irrational.

For \(ax^2+bx+c=0\), two distinct real roots require the discriminant \(D=b^2-4ac>0\). In option C, \(D=(-5)^2-4\cdot2\cdot2=9>0\). Option A has \(D=0\), so its roots are equal. Exam tip: check the sign of \(D\) first.

A quadratic equation has equal roots only when its discriminant is D = 0. Hence, (s − 2)^2 = 0, which gives s − 2 = 0 and therefore s = 2. If s = −2, the discriminant becomes 16, indicating two distinct real roots. Exam tip: remember that equal roots require D = 0.

A quadratic equation has no real roots when its discriminant satisfies \(D<0\). Therefore, \((u+1)(u-5)<0\). The zeros of the two factors are \(-1\) and \(5\), and their product is negative between these values, giving \(-1<u<5\). At the endpoints, \(D=0\), so the equation has two equal real roots. Exam tip: First write the required discriminant condition, then check the sign of the product between its critical values.

A quadratic equation has real roots when its discriminant satisfies \(D\geq 0\). Thus, \(8m-24\geq 0\), which gives \(8m\geq 24\) and hence \(m\geq 3\). At \(m=3\), the roots are equal; for \(m>3\), the roots are distinct and real. Exam tip: when solving a discriminant inequality, check whether dividing by a negative quantity would reverse the inequality sign.

Here, \(a=1\), \(b=2(1-\sqrt{3})\), and \(c=4\). Therefore, the discriminant is \(D=b^2-4ac=4(1-\sqrt{3})^2-16=-8\sqrt{3}<0\). Hence, the equation has no real roots. Option B would be correct only if \(D=0\). Exam tip: For a quadratic equation, \(D<0\) indicates non-real, or complex, roots.

Here, \\(a=1\\), \\(b=-2(3+\sqrt{2})\\), and \\(c=17+12\sqrt{2}\\). Therefore, the discriminant is \\(D=b^2-4ac=4(3+\sqrt{2})^2-4(17+12\sqrt{2})=-24(1+\sqrt{2})<0\\). Hence, the quadratic has no real roots; its roots are a pair of complex conjugates. Remember that real and equal roots occur only when \\(D=0\\).

The discriminant of the quadratic is \(D=b^2-4ac=(-2r)^2-4(1)(r^2+9)=-36\). It remains negative for every real value of \(r\), so the equation has no real roots. In fact, its roots are \(r\pm3i\), which are complex and non-real. Exam tip: If \(D<0\), a quadratic equation has no real roots.

On the \(x\)-axis, \(y=0\), so we get \(x^2-2kx+k^2+1=0\). Its discriminant is \(D=(-2k)^2-4(1)(k^2+1)=-4<0\), so it has no real roots for any real \(k\). Hence, the parabola has no real intersection with the \(x\)-axis. Equivalently, \(y=(x-k)^2+1\ge 1\), so its vertex always lies one unit above the \(x\)-axis. Exam tip: For a quadratic graph, \(D<0\) means there are no real intersections with the \(x\)-axis.

The equation can be written as \(y=x^2-2kx+k^2=(x-k)^2\). On the x-axis, \(y=0\), so \((x-k)^2=0\), giving the repeated root \(x=k\). Therefore, the parabola touches the x-axis at exactly one point, \((k,0)\), for every real value of \(k\). It does not cut the axis at two distinct points because that would require \(D>0\), whereas here \(D=0\). Exam tip: for a quadratic, \(D=0\) indicates equal roots and tangency to the x-axis.

On the \(x\)-axis, \(y=0\). Therefore, we solve \(x^2-2kx+k^2-4=0\). Its discriminant is \(\Delta=(-2k)^2-4(1)(k^2-4)=16>0\), so for every real value of \(k\), the equation has two distinct real roots. In fact, the roots are \(x=k-2\) and \(x=k+2\), so the parabola intersects the \(x\)-axis at \((k-2,0)\) and \((k+2,0)\). Remember: \(\Delta>0\) indicates two distinct points of intersection.

Here, \(a=1\), \(b=-2p\), and \(c=p^2-5p\). Therefore, the discriminant is \(D=b^2-4ac=(-2p)^2-4(p^2-5p)=20p\). A quadratic equation has two real and distinct roots only when \(D>0\). Hence, \(20p>0\), which gives \(p>0\). For \(p=0\), the discriminant is zero and the roots are equal; for \(p<0\), the roots are non-real. Exam tip: For distinct real roots, always apply the condition \(D>0\).

Here \\(A=1\\), \\(B=-(2a+1)\\), and \\(C=a(a+1)\\). Therefore, the discriminant is \\(D=B^2-4AC=(2a+1)^2-4a(a+1)=1\\). Since \\(D>0\\), the roots are real and distinct. In fact, the roots are \\(a\\) and \\(a+1\\). They cannot always be called rational or irrational, because that depends on the value of \\(a\\). Exam tip: simplify the discriminant first and then determine its sign.

The equation can be factorised as \(x^2-(a+b)x+ab=(x-a)(x-b)=0\). Hence, its roots are \(x=a\) and \(x=b\). Since a and b are real and unequal, the roots are distinct and real. Equal roots would occur only when \(a=b\), and the roots need not be irrational. Exam tip: Use the discriminant \(D=(a-b)^2\); when \(a\neq b\), \(D>0\), which indicates two distinct real roots.

A quadratic equation has no real roots when its discriminant satisfies \(D<0\). Here, \(D=[-2(a-1)]^2-4(a^2+1)=4(a-1)^2-4(a^2+1)=-8a\). Therefore, \(-8a<0\), which gives \(a>0\). For \(a=0\), \(D=0\), so the equation has two real and equal roots. Exam tip: whenever ‘no real roots’ is stated, immediately apply the condition \(D<0\).

The direct answer is option A: the graph does not cut the x-axis and there are no real roots. For a quadratic equation ax^2+bx+c=0 with a≠0, the discriminant is D=b^2−4ac. The quadratic formula is x=(-b±√D)/(2a). If D<0, its square root is not a real number, so the equation has no real solutions. On the graph y=ax^2+bx+c, an x-intercept is exactly a real value of x for which y=0. Since no real solution exists, the parabola has no x-intercept and does not meet or cross the x-axis. Option A is correct. Option B describes D=0, when the graph touches the axis once and the roots are equal. Option C describes D>0, when there are two distinct real roots and two intersections. Option D is false because a negative discriminant does not make roots rational; it makes them non-real. A quick memory rule is D<0: no real roots, no x-axis intersection.

Here, \(a=2\), \(b=-6\sqrt{2}\), and \(c=9\). Therefore, the discriminant is \(\Delta=b^2-4ac=(-6\sqrt{2})^2-4(2)(9)=72-72=0\). When \(\Delta=0\), the quadratic equation has two real and equal roots. In fact, the repeated root is \(x=\frac{3\sqrt{2}}{2}\). Hence, option A is correct; options B and D require a positive discriminant, whereas the discriminant here is zero. Exam tip: To determine the nature of roots, first check the sign of \(\Delta\).

Here, \(a=3\), \(b=-4\sqrt{3}\), and \(c=5\). Therefore, the discriminant is \(\Delta=b^2-4ac=(-4\sqrt{3})^2-4(3)(5)=48-60=-12\). Since \(\Delta<0\), the equation has no real roots. Option B is incorrect because equal real roots require \(\Delta=0\). Exam tip: For a quadratic equation, a negative discriminant always indicates that no real roots exist.

Here, \\(a=5\\), \\(b=-2\\sqrt{30}\\), and \\(c=6\\). Therefore, the discriminant is \\(\\Delta=b^2-4ac=(-2\\sqrt{30})^2-4(5)(6)=120-120=0\\). Hence, the roots are real and equal. Option D is incorrect because \\(\\Delta>0\\) would indicate distinct roots. Exam tip: For a quadratic equation, \\(\\Delta=0\\) always means two equal real roots.

For a quadratic equation to have real roots, its discriminant must satisfy \(D\geq 0\). Here, \(D=[-2(k+2)]^2-4(k^2+3k+7)=4(k-3)\). Therefore, \(4(k-3)\geq 0\), which gives \(k\geq 3\). Option B is incorrect because \(k=3\) also gives real, equal roots. Exam tip: For questions about real roots, begin by applying \(D\geq 0\).