Nature of Roots

When the discriminant \(D=b^2-4ac\) is zero, \(\sqrt{D}=0\) in \(\frac{-b\pm\sqrt{D}}{2a}\), so both roots are equal. For \(D>0\), the roots are distinct. Exam tip: equal roots always mean \(D=0\).

Here, \(a=1\), \(b=-10\), and \(c=25\). Therefore, the discriminant is \(D=b^2-4ac=(-10)^2-4(1)(25)=100-100=0\). For a quadratic equation, \(D=0\) means that the two roots are equal. In fact, \(x^2-10x+25=(x-5)^2\), so both roots are \(5\). Hence, both the assertion and the reason are correct, and the reason correctly explains the assertion. Exam tip: Calculate \(D\) first to determine the nature of the roots.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=1\), \(b=2\), and \(c=3\), so \(D=2^2-4(1)(3)=4-12=-8\). The value 16 results from using the incorrect formula \(b^2+4ac\). Exam tip: remember the discriminant formula as \(b^2-4ac\).

Here, a = 9, b = 12, and c = 4. Therefore, the discriminant is D = b² − 4ac = 12² − 4 × 9 × 4 = 144 − 144 = 0. Hence, the roots are real and equal. In fact, 9x² + 12x + 4 = (3x + 2)², so the repeated root is x = −2/3. Option B would require D > 0, but the discriminant here is zero. Exam tip: For a quadratic equation, D = 0 indicates two equal real roots.

Here, \(a=7\), \(b=-2\), and \(c=5\). Therefore, the discriminant is \(D=b^2-4ac=(-2)^2-4(7)(5)=4-140=-136\). Since \(D<0\), the equation has no real roots. Option C results from taking the sign of the discriminant incorrectly. In an exam, calculate \(D\) first and then check its sign to determine the nature of the roots.

Here, \(a=2\), \(b=-7\), and \(c=3\). Therefore, the discriminant is \(D=b^2-4ac=(-7)^2-4(2)(3)=25\). Since \(D>0\), the roots are real and distinct; since \(D=25\) is a perfect square, they are also rational. In fact, the roots are \(3\) and \(\frac{1}{2}\). Hence, option A is correct. Exam tip: \(D>0\) indicates distinct real roots, while a perfect-square discriminant indicates rational roots.

Here, \(a=6, b=-1, c=-2\), so the discriminant is \(D=b^2-4ac=(-1)^2-4(6)(-2)=49\). Since \(D\) is positive and a perfect square, the roots are real, rational, and distinct. Option B would require \(D=0\), while the value \(D=7\) in option D is incorrect. Exam tip: If \(D>0\) and is a perfect square, the roots are rational and distinct.

For a quadratic equation \\(ax^2+bx+c=0\\), the discriminant is \\(D=b^2-4ac\\). Here, \\(a=4,b=4,c=5\\), so \\(D=4^2-4(4)(5)=16-80=-64\\). Since \\(D<0\\), the equation has no real roots. The roots are real and equal only when \\(D=0\\), so option B is not correct. Exam tip: determine the nature of roots by checking the sign of the discriminant first.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=1, b=7, c=10\), so \(D=7^2-4(1)(10)=49-40=9\). Since \(D\) is positive and a perfect square, the roots are real, rational, and distinct; in fact, they are \(-5\) and \(-2\). Therefore, option A is correct. Exam tip: \(D>0\) gives distinct real roots, and a perfect-square discriminant makes those roots rational.

Here, \(a=3\), \(b=-6\), and \(c=3\). Therefore, the discriminant is \(D=b^2-4ac=(-6)^2-4(3)(3)=36-36=0\). Hence, the roots are real and equal. In fact, the equation can be written as \(3(x-1)^2=0\), giving the repeated root \(x=1\). Option B is incorrect because its discriminant value is wrong. Exam tip: When \(D=0\), the roots are always real and equal.

A quadratic equation \(ax^2+bx+c=0\) has two real and equal roots when its discriminant \(D=b^2-4ac\) is zero. For option (A), \(a=1, b=-6, c=9\), so \(D=(-6)^2-4(1)(9)=36-36=0\). Hence its roots are equal, namely \(x=3,3\). Options (B) and (D) have discriminants 4 and 16, respectively, so their roots are distinct; option (C) has discriminant \(-4\), so its roots are not real. Exam tip: For equal roots, check directly whether \(D=0\).

For a quadratic equation ax² + bx + c = 0, the discriminant is D = b² − 4ac. In option (A), D = 2² − 4(1)(5) = −16. Since D < 0, the equation has no real roots. In option (D), D = 0, so it has one repeated real root, while options (B) and (C) have D > 0. Exam tip: Check the sign of the discriminant first to determine the nature of the roots.

For a quadratic equation 8ax^2+bx+c=09 to have equal roots, its discriminant 8D=b^2-4ac9 must be zero. Here, 8a=1,b=p,c=169, so 8D=p^2-4\times1\times16=p^2-64=09. Hence 8p^2=649 and 8p=\pm89, making option A correct. For the closest distractor, 8p=\pm49 gives a non-zero discriminant. Exam tip: whenever roots are equal, set the discriminant equal to zero first.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=1, b=-4, c=1\), so \(D=(-4)^2-4(1)(1)=12\). Since \(D>0\) and 12 is not a perfect square, the roots are real, irrational, and distinct. Option B is incorrect because \(D=16\) is not the discriminant of this equation. Exam tip: First use the sign of \(D\) to determine whether the roots are real and distinct, then check whether \(D\) is a perfect square to determine rationality.

For a quadratic equation \(ax^2+bx+c=0\), the roots are equal only when the discriminant \(D=b^2-4ac\) is zero. Therefore, the equation with \(D_1=0\) has equal roots. When \(D_2=7>0\), the roots are distinct and real, while \(D_3=-2<0\) gives no real roots. Exam tip: For equal roots, check \(D=0\) directly.

Here, \(a=1\), \(b=-8\), and \(c=16\). Thus, the discriminant is \(D=b^2-4ac=(-8)^2-4(1)(16)=64-64=0\). When \(D=0\), a quadratic equation has two equal real roots. In fact, \(t^2-8t+16=(t-4)^2\), so both roots are \(t=4\). Option B applies only when \(D>0\), which gives distinct real roots. Exam tip: To determine the nature of roots, first check the sign of the discriminant.

For equal roots, the discriminant \\(D=b^2-4ac\\) must be zero. Here, \\(a=q\\), \\(b=5\\), and \\(c=q\\), so \\(D=25-4q^2=0\\). Thus, \\(q^2=\\frac{25}{4}\\), giving \\(q=\\pm\\frac{5}{2}\\). The values \\(5\\) and \\(-5\\) in option B do not make the discriminant zero. Also, \\(q=0\\) is excluded because it would make the equation non-quadratic. Exam tip: for equal roots, set the discriminant directly equal to zero.

A quadratic equation has no real roots when its discriminant satisfies \(D<0\). Here, \(a=1\), \(b=m-2\), and \(c=1\), so \(D=(m-2)^2-4\). Thus, \((m-2)^2<4\), which gives \(-2<m-2<2\), and hence \(0<m<4\). Therefore, option A is correct. In option C, \(D=0\), which gives two equal real roots rather than no real roots. Exam tip: For ‘no real roots’, first apply the condition \(D<0\).

For equal roots, the discriminant \(D=b^2-4ac\) must be zero. Here, \(a=1\), \(b=-2(k-1)\), and \(c=k+1\). Thus, \(D=4(k-1)^2-4(k+1)=4(k^2-3k)=4k(k-3)\). Setting \(D=0\) gives \(k=0\) or \(k=3\), so option A is correct. Exam tip: whenever a quadratic equation has equal roots, begin by applying \(D=0\).

A quadratic equation has no real roots when its discriminant satisfies \(D<0\). Here, \(a=2\), \(b=k-4\), and \(c=8\), so \(D=(k-4)^2-4(2)(8)=(k-4)^2-64\). Therefore, \((k-4)^2<64\), which gives \(-8<k-4<8\), and hence \(-4<k<12\). Thus, option A is correct. In option B, \(D>0\), giving two distinct real roots, while in option C, \(D=0\), giving equal real roots. Exam tip: For the nature of roots, first calculate the discriminant using \(D=b^2-4ac\).