Nature of Roots

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=5\), \(b=-20\), and \(c=20\), so \(D=(-20)^2-4(5)(20)=400-400=0\). Therefore, the correct answer is 0. When \(D=0\), the two roots are equal. Exam tip: Always substitute the signed value of \(b\) before squaring it.

Here, \(a=5, b=-20, c=20\). The discriminant is \(D=b^2-4ac=(-20)^2-4(5)(20)=400-400=0\). Therefore, the roots are real and equal. In fact, the equation becomes \(5(x-2)^2=0\), so both roots are \(x=2\). Exam tip: \(D=0\) gives real and equal roots, whereas \(D>0\) gives distinct real roots.

Here, \(D=25>0\), so the two roots are real and distinct. Since 25 is a perfect square and the coefficients are rational, the roots are also rational. Option B would be correct only when \(D=0\), while irrational roots occur when the positive discriminant is not a perfect square. Exam tip: \(D>0\) gives distinct real roots, \(D=0\) gives equal real roots, and \(D<0\) gives non-real roots.

For a quadratic equation, D>0 means that the roots are real and distinct. Here, D=5>0, so the roots are real and distinct. Since 5 is not a perfect square and the coefficients are integers, the roots are also irrational. Therefore, option A is correct. Exam tip: D>0 gives real and distinct roots, D=0 gives equal roots, and D<0 gives non-real roots.

For a quadratic equation, the discriminant is \(D=b^2-4ac\). If \(D<0\), the equation has no real roots. Here, \(D=-9<0\), so option A is correct. Option B requires \(D>0\), while option C requires \(D=0\). Exam tip: determine the nature of the roots by checking the sign of the discriminant.

For equal roots, the discriminant must be zero. Here, \(a=1\), \(b=k\), and \(c=16\), so \(D=b^2-4ac=k^2-4(1)(16)=k^2-64\). Thus, \(k^2-64=0\), giving \(k^2=64\). Exam tip: For equal roots, immediately apply \(b^2-4ac=0\); do not confuse the constant term 16 with the required value of \(k^2\).

For \(ax^2+bx+c=0\), the product of the roots is \(c/a\). In option C, \(c/a=-6\), so one root is positive and the other is negative. Option A has a positive product, so its roots cannot have opposite signs. Exam tip: check whether \(c/a<0\).

For a quadratic equation \(ax^2+bx+c=0\), the roots are real and distinct when the discriminant \(D=b^2-4ac\) is positive. Here, \(a=1\), \(b=p\), and \(c=6\), so \(D=p^2-24\). Therefore, \(p^2-24>0\), which gives \(p^2>24\). If \(p^2=24\), the roots are real but equal. Exam tip: remember that real and distinct roots require \(D>0\).

A quadratic equation has no real roots when its discriminant \(D=b^2-4ac\) is negative. Here, \(a=1\), \(b=q\), and \(c=10\), so \(D=q^2-40\). Therefore, \(q^2-40<0\), which gives \(q^2<40\). Remember that \(q^2=40\) gives equal real roots, not no real roots.

For equal roots, the discriminant \(D=b^2-4ac\) must be zero. Here, \(a=m\), \(b=6\), and \(c=3\), so \(6^2-4(m)(3)=0\). Thus, \(36-12m=0\), giving \(m=3\). Therefore, option A is correct. Exam tip: For equal-root questions, immediately apply the condition \(D=0\).

For equal roots, the discriminant \(D=b^2-4ac\) must be zero. Here, \(a=2\), \(b=n\), and \(c=8\), so \(n^2-4(2)(8)=0\), giving \(n^2=64\). Remember that the question asks for \(n^2\), not \(n\).

For option A, the discriminant is \(D=b^2-4ac=(-9)^2-4(1)(20)=1\). Since \(D>0\) and 1 is a perfect square, the roots are real, rational, and distinct. In fact, the equation factors as \((x-4)(x-5)=0\), giving roots 4 and 5. The discriminants of options B, C, and D are \(-3,-39,-16\), respectively, so their roots are not real. Exam tip: For a quadratic equation, \(D>0\) and a perfect-square discriminant indicate distinct real rational roots.

For option B, \(a=1, b=-2, c=-2\). Therefore, the discriminant is \(\Delta=b^2-4ac=(-2)^2-4(1)(-2)=12\). Since \(\Delta>0\), the roots are real and distinct; since 12 is not a perfect square, the roots are irrational. In fact, the roots are \(1\pm\sqrt{3}\). Option A has discriminant 16, giving rational roots; option C has discriminant 0, giving equal roots; and option D has a negative discriminant, so it has no real roots. Exam tip: real, distinct, irrational roots require \(\Delta>0\) and \(\Delta\) not to be a perfect square.

The direct answer is A: the statement is true. For a quadratic equation written as ax^2+bx+c=0, where a is not zero, the discriminant is D=b^2-4ac. The roots are given by x=(-b plus or minus square root of D)/(2a). When D=0, square root of D is 0, so both signs give the same value, x=-b/(2a). Therefore the two roots are real and equal. Option A is correct. Option B is wrong because the statement is a standard true rule. Option C is wrong because a=0 would remove the quadratic term, so the equation would not be a quadratic equation. Option D is wrong because c=0 is not required; it only makes one root zero in some cases. Exam cue: D=0 means equal roots.

For a quadratic equation, the discriminant is \(D=b^2-4ac\). When \(D<0\), the equation has no real roots, so the statement claiming two real and distinct roots is false. Two real and distinct roots occur when \(D>0\), while \(D=0\) gives equal real roots. Exam tip: Always check the three cases \(D>0\), \(D=0\), and \(D<0\) to determine the nature of the roots.

The statement is incorrect. For a quadratic equation, \(D>0\) gives two real and distinct roots, not equal roots. Equal roots occur only when \(D=0\), while \(D<0\) gives no real roots. Exam tip: remember the sequence—\(D>0\): real and distinct roots, \(D=0\): equal roots, \(D<0\): non-real roots.

Here, \(a=4\), \(b=-4\), and \(c=3\). The discriminant is \(\Delta=b^2-4ac=(-4)^2-4(4)(3)=16-48=-32<0\). Therefore, the equation has no real roots. Option D is incorrect because irrational and distinct roots are real roots and require a positive discriminant. Exam tip: For a quadratic equation, \(\Delta<0\) means that it has no real roots.

In the standard form \(ax^2+bx+c=0\), \(a=6\), \(b=-5\), and \(c=1\). Thus, \(D=b^2-4ac=(-5)^2-4(6)(1)=25-24=1\), so option A is correct. Since \(D>0\), the roots are real and distinct; because \(D\) is also a perfect square, they are rational. Exam tip: always place the signed value of \(b\) in parentheses when calculating the discriminant.

Here, the discriminant is \(D=b^2-4ac=(-5)^2-4(6)(1)=25-24=1\). Since \(D\) is positive and a perfect square, the roots are real, rational, and distinct. In fact, the roots are \(x=\frac{1}{2}\) and \(x=\frac{1}{3}\). Exam tip: if \(D>0\) is a perfect square, the roots are real, rational, and distinct.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=3\), \(b=-6\), and \(c=1\), so \(D=(-6)^2-4(3)(1)=36-12=24\). Therefore, option A is correct. Remember that a positive discriminant indicates two distinct real roots; 36 is only the value of \(b^2\), not the complete discriminant.

The discriminant of ax² + bx + c = 0 is D = b² − 4ac. In this equation, a = 3, b = −6 and c = 1. Thus D = (−6)² − 4(3)(1) = 36 − 12 = 24. A positive discriminant shows that the two roots are real and distinct. To decide whether they are rational or irrational, observe that 24 is not a perfect square. Indeed, the quadratic formula gives x = [6 ± √24]/6 = 1 ± √6/3, and √6 is irrational. Hence both roots are real, irrational and distinct. Therefore option A is correct. Option B would require a positive perfect-square discriminant, while options C and D correspond to D = 0 and D < 0 respectively.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=1, b=8, c=16\), so \(D=8^2-4(1)(16)=64-64=0\). Therefore, the equation has equal roots. In an exam, identify the signs of \(a,b,c\) before substitution; \(64\) is only the value of \(b^2\), not the complete discriminant.

Here, \(a=1, b=8, c=16\), so the discriminant is \(D=b^2-4ac=8^2-4(1)(16)=0\). Therefore, the roots are real and equal. In fact, \(x^2+8x+16=(x+4)^2\), so both roots are \(x=-4\). Exam tip: when \(D=0\), a quadratic equation has real and equal roots.

Comparing the equation with \(ax^2+bx+c=0\), we get \(a=2, b=7, c=3\). Thus, \(D=b^2-4ac=7^2-4(2)(3)=49-24=25\). The distractor 49 is only \(b^2\), without subtracting \(4ac\). Exam tip: Calculate the value and sign of \(D\) first to determine the nature of the roots.

Here, a = 2, b = 7 and c = 3. The discriminant is D = b² − 4ac = 7² − 4(2)(3) = 25. Since D > 0, the roots are real and distinct; since 25 is a perfect square, they are also rational. In fact, the roots are −1/2 and −3. Therefore, option A is correct. Exam tip: If D > 0 and D is a perfect square, the roots are real, rational and distinct.