Nature of Roots

The governing concept is the discriminant D = b² − 4ac, which determines the nature of quadratic roots. For 2x² − 3√2 x + 7 = 0, we have a = 2, b = −3√2, and c = 7. Thus b² = (−3√2)² = 18, while 4ac = 4 × 2 × 7 = 56. Therefore D = 18 − 56 = −38. Since the discriminant is negative, the equation has no real roots; its two roots are complex conjugates. Hence option A is correct. Option B would require D = 0, and options C and D incorrectly assert positive discriminants and real roots. The irrational coefficient does not change the discriminant rule; it must simply be squared accurately.

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.

For a quadratic equation ax² + bx + c = 0, the discriminant D = b² − 4ac determines the nature of its roots. Here a = 1, b = −2p and c = p² − 25. Therefore, D = (−2p)² − 4(1)(p² − 25) = 4p² − 4p² + 100 = 100. This is positive, so the roots are real and distinct. It is also a perfect square, and the quadratic formula gives x = [2p ± 10]/2 = p ± 5, which are rational whenever p is rational; in the intended school classification, the constant square-root part confirms rational distinct roots. Hence option A is correct. Options B and C would require D = 0 and D < 0 respectively, while D being non-square would lead to irrational roots.

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.

For a quadratic equation ax² + bx + c = 0, the discriminant is D = b² − 4ac. Compare x² − 2(k−2)x + k² = 0 with the standard form. Here a = 1, b = −2(k−2), and c = k². Therefore D = [−2(k−2)]² − 4(1)(k²) = 4(k−2)² − 4k². Expanding gives 4(k² − 4k + 4) − 4k² = 4k² − 16k + 16 − 4k² = 16 − 16k = 16(1−k). Hence option A is correct. Option B results from incorrectly discarding the k-dependent terms. Option C has an incorrect sign and expression, while option D is not obtained from b² − 4ac. Because k is a parameter, the correct discriminant remains an algebraic expression in k.

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.