Nature of Roots

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=2, b=5, c=2\), so \(D=5^2-4(2)(2)=25-16=9\). Therefore, the correct answer is 9. Since \(D>0\), the equation has two distinct real roots. In exams, carefully retain the minus sign in \(b^2-4ac\); using a plus sign would incorrectly give 41.

Here (D=9>0), so there will be two real and distinct roots. Finding (D) first is the safest method.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=3\), \(b=2\), and \(c=1\), so \(D=(2)^2-4(3)(1)=4-12=-8\). Therefore, option A is correct. Choosing 8 usually results from missing the minus sign in \(-4ac\). Exam tip: identify \(a\), \(b\), and \(c\) first, then substitute them carefully.

Here, \(a=3, b=2, c=1\). The discriminant is \(D=b^2-4ac=2^2-4(3)(1)=4-12=-8\). Since \(D<0\), the equation has no real roots; its roots are complex. Roots are real and distinct when \(D>0\), and real and equal when \(D=0\). In an exam, check the sign of the discriminant first to determine the nature of the roots.

For the given quadratic equation, \(a=1\), \(b=6\), and \(c=9\). Thus, the discriminant is \(D=b^2-4ac=6^2-4(1)(9)=36-36=0\). When \(D=0\), the two roots are real and equal; here, both roots are \(x=-3\). Therefore, option A is correct. Exam tip: \(D=0\) indicates equal roots, \(D>0\) indicates distinct real roots, and \(D<0\) indicates no real roots.

Here, \\(a=1, b=7, c=10\\). Therefore, \\(D=b^2-4ac=7^2-4(1)(10)=49-40=9\\). Hence, the correct answer is 9. The value 49 is only \\(b^2\\), not the complete discriminant. Exam tip: identify \\(a,b,c\\) first and substitute them carefully into \\(b^2-4ac\\).

For a quadratic equation ax^2+bx+c=0, the discriminant D=b^2−4ac determines the nature of its roots. Here a=1, b=7 and c=10, so D=7^2−4(1)(10)=49−40=9. Since D is positive, the equation has two real and distinct roots. In fact, factorisation gives x^2+7x+10=(x+5)(x+2), so the roots are −5 and −2, which confirms that they are real and unequal. Option B would require D=0, while option C applies when D<0. A quadratic with real coefficients cannot have only one isolated imaginary root; non-real roots occur as a conjugate pair.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=1\), \(b=4\), and \(c=8\), so \(D=4^2-4(1)(8)=16-32=-16\). Therefore, the correct answer is -16. Choosing 16 results from a sign error. Exam tip: when \(D<0\), the roots are not real.

Here, \(a=1\), \(b=4\), and \(c=8\). The discriminant is \(D=b^2-4ac=4^2-4(1)(8)=16-32=-16<0\). Therefore, the equation has no real roots. The roots would be real and equal only if \(D=0\). In an exam, first check the sign of the discriminant to determine the nature of the roots.

For a quadratic equation \(ax^2+bx+c=0\), equal roots require the discriminant \(D=b^2-4ac\) to be zero. Here, \(a=1\), \(b=k\), and \(c=9\), so \(D=k^2-4(1)(9)=k^2-36\). Therefore, \(k^2-36=0\), giving \(k^2=36\). The other options result from using an incorrect value for \(4ac\) or from ignoring the discriminant condition. Exam tip: for equal roots, immediately set \(D=0\).

For equal roots, the discriminant must be zero. Thus, \(k^2-4(1)(9)=0\), giving \(k^2-36=0\) and hence \(k=6\) or \(k=-6\). Among the given options, only 6 is listed, so option A is correct. Exam tip: For equal roots of a quadratic equation, always set \(b^2-4ac=0\).

For equal roots, the discriminant must be \(\Delta=0\). Here, \(a=1\), \(b=-2k\), and \(c=16\), so \(\Delta=b^2-4ac=(-2k)^2-4(1)(16)=0\). Thus, \(4k^2-64=0\), giving \(k^2=16\). The value 8 is incorrect because simplifying the equation still leads to \(k^2=16\). Exam tip: For equal roots of a quadratic equation, set \(b^2-4ac=0\).

The discriminant is \(D=b^2-4ac\). If \(D<0\), its square root is not real, so the equation has non-real roots. In contrast, \(D=0\) gives equal real roots. Exam tip: determine the nature of roots by checking the sign of \(D\).

For a quadratic equation \(ax^2+bx+c=0\) to have equal roots, its discriminant must be zero: \(D=b^2-4ac=0\). Here, \(a=k\), \(b=4\), and \(c=1\). Thus, \(4^2-4(k)(1)=0\), giving \(16-4k=0\) and hence \(k=4\). Therefore, option A is correct. Exam tip: For equal roots, immediately apply the condition \(D=0\).

For a quadratic equation to have equal roots, its discriminant must be zero. Here, \(a=2\), \(b=k\), and \(c=2\), so \(D=b^2-4ac=k^2-4(2)(2)=k^2-16\). Thus, \(k^2-16=0\), giving \(k^2=16\). Exam tip: For equal-root questions, immediately use the condition \(b^2-4ac=0\).

For a quadratic equation ax² + bx + c = 0, the discriminant is Δ = b² − 4ac. In option A, Δ = (−3)² − 4(1)(2) = 1 > 0, so its roots are real and distinct. Options B and D have Δ = 0, giving equal roots, while option C has Δ = −3 < 0, so its roots are not real. Exam tip: Δ > 0 indicates real and distinct roots.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). In option A, \(D=(-8)^2-4(1)(16)=64-64=0\), so its roots are real and equal. Options B and D have \(D>0\), giving two distinct real roots, while option C has \(D<0\), so it has no real roots. Exam tip: equal real roots occur exactly when \(D=0\).

For a quadratic equation \(ax^2+bx+c=0\), the nature of the roots is determined by the discriminant \(D=b^2-4ac\). For option A, \(D=2^2-4(1)(10)=-36<0\), so it has no real roots. Option B has \(D=0\), while options C and D have \(D>0\), so they have real roots. Exam tip: roots are non-real when \(D<0\).

Here, \(a=1, b=-10, c=25\). Therefore, the discriminant is \(D=b^2-4ac=(-10)^2-4(1)(25)=100-100=0\). Hence the roots are real and equal; in fact, the equation becomes \((x-5)^2=0\), giving the repeated root \(x=5\). Option B is incorrect because distinct real roots require \(D>0\). Exam tip: In a quadratic equation, \(D=0\) indicates equal roots immediately.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=2, b=-7, c=3\), so \(D=(-7)^2-4(2)(3)=49-24=25>0\). Therefore, the roots are real and distinct. Option B would apply if \(D=0\), while option C would require \(D<0\). In an exam, first check the sign of the discriminant to determine the nature of the roots.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(D=1^2-4(6)(1)=-23<0\), so the equation has no real roots. Option C would apply only if \(D=0\), but the discriminant here is negative. Exam tip: when \(D<0\), the roots are non-real.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=7\), \(b=-10\), and \(c=3\), so \(D=(-10)^2-4(7)(3)=100-84=16\). Therefore, the correct answer is 16. The distractor 100 results from calculating only \(b^2\) and omitting \(4ac\). Exam tip: always place a negative value of \(b\) in parentheses before squaring it.

The standard test for the nature of roots is the discriminant D=b^2−4ac. Comparing 7x^2−10x+3=0 with ax^2+bx+c=0 gives a=7, b=−10 and c=3. Hence D=(−10)^2−4(7)(3)=100−84=16. Because D>0, the equation has two real and distinct roots, so option A is correct. We can also verify this by factorisation: 7x^2−10x+3=(7x−3)(x−1), giving roots 3/7 and 1, which are real and unequal. Equal real roots would require D=0, and non-real roots would require D<0. The roots are also not both negative, so option D is contradicted by their actual positive values.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=8\), \(b=8\), and \(c=2\), so \(D=8^2-4(8)(2)=64-64=0\). Therefore, the correct answer is 0. The value 64 is only \(b^2\), not the complete discriminant. Exam tip: when \(D=0\), the two roots are equal.

The nature of the roots of a quadratic equation is determined by its discriminant. Here, D = 4, so D > 0; therefore, the roots are real and distinct. Option B is incorrect because equal real roots occur when D = 0. Exam tip: remember D > 0, D = 0 and D < 0 as distinct real, equal real and non-real roots, respectively.