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 \(q\ne0\), \(D<0\), 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.

The x-intercepts of the parabola y=x²+2x+5 are obtained by solving x²+2x+5=0. Its discriminant is D=b²−4ac=2²−4(1)(5)=4−20=−16. A negative discriminant means the quadratic has no real zeros. Consequently, the graph has no point whose y-coordinate is zero, so it does not meet or cut the x-axis. The assertion is therefore true, and the stated reason is also true and directly explains it. Hence option A is correct. If D were zero, the graph would touch the x-axis once; if D were positive, it would cut the axis at two points. Options B, C, and D incorrectly reject either the assertion, the calculation, or both.

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.

The governing concept is the discriminant condition for coincident roots. Equal real roots occur when D=0. Since D=25−4q², impose the condition 25−4q²=0. Rearranging gives 4q²=25, so q²=25/4. Taking square roots gives q=±√(25/4)=±5/2; both signs must be included because a positive and a negative number have the same square. Thus option A is correct. Option B results from the error q²=25, which ignores the factor 4. For q=±2, D=25−16=9, not zero, so the roots are distinct real roots. For q=0, D=25, also not zero. These checks rule out the remaining choices.

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\\).

For a quadratic ax² + bx + c = 0, the discriminant is D = b² - 4ac. In this equation, a = 1, b = -2r, and c = r² - 16. Therefore D = (-2r)² - 4(1)(r² - 16) = 4r² - 4r² + 64 = 64. This value is positive for every real r, so the roots are always real and distinct. Moreover, 64 is a perfect square, so the quadratic formula produces rational roots: x = (2r ± 8)/2 = r ± 4. Hence option A is correct. Equal roots would require D = 0, no real roots require D < 0, and irrational roots would require a positive but non-square discriminant.

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.

A real value of l can occur only when the quadratic equation has real roots, so its discriminant must satisfy D ≥ 0. Comparing l² - 2(a + 3)l + (a² + 10) with Al² + Bl + C, we have A = 1, B = -2(a + 3), and C = a² + 10. Thus D = B² - 4AC = 4(a + 3)² - 4(a² + 10) = 4[(a² + 6a + 9) - a² - 10] = 4(6a - 1). The condition 4(6a - 1) ≥ 0 gives a ≥ 1/6. Therefore option A is correct. The other choices either reverse the inequality, impose an unnecessary single value, or ignore the discriminant condition.

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 \(D>0\).