Nature of Roots

For a quadratic equation, the discriminant is \\(D=b^2-4ac\\). Here, \\(D=9>0\\), so the two roots are real and distinct. Equal real roots require \\(D=0\\), so option B is incorrect. Exam tip: \\(D>0\\) indicates real and distinct roots.

For a quadratic equation, the discriminant is \(D=b^2-4ac\). Here, \(D=-7<0\), so the equation has no real roots; its roots are complex. Option B requires \(D>0\), while option C requires \(D=0\). Exam tip: whenever \(D<0\), conclude that there are no real roots.

When (D=0), both roots are equal, so there is (1) different real root. Remember the difference between equal and distinct.

For a quadratic equation \\(ax^2+bx+c=0\\), the discriminant is \\(D=b^2-4ac\\). Here, \\(a=4, b=4, c=1\\), so \\(D=4^2-4(4)(1)=16-16=0\\). Therefore, the correct answer is 0, and the roots are equal. Choosing 8 or 16 results from an incorrect subtraction in the formula. Exam tip: when \\(D=0\\), the quadratic equation has two equal real roots.

Here, \(a=4, b=4, c=1\). The discriminant is \(D=b^2-4ac=4^2-4(4)(1)=16-16=0\). Therefore, the roots are real and equal; in fact, each root is \(-\frac{1}{2}\). Option B would require \(D>0\), but here \(D=0\). Exam tip: For a quadratic equation, \(D=0\) always indicates real and equal 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 \(D=(-2)^2-4(1)(-3)=4+12=16\). Since \(c\) is negative, the term \(-4ac\) becomes positive; therefore, 8 and -16 are incorrect. In an exam, identify the signs of \(a\), \(b\), and \(c\) before substituting.

For ax²+bx+c=0, calculate the discriminant D=b²−4ac. In this equation, a=1, b=−2, and c=−3, so D=(−2)²−4(1)(−3)=4+12=16. Because D>0, the quadratic has two real and distinct roots. Indeed, x²−2x−3 factors as (x−3)(x+1), giving roots 3 and −1, which are real and unequal. Thus option A is correct. Equal real roots occur only when D=0, and non-real roots occur when D<0; option D is also inconsistent because the actual roots are not equal and one is positive.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=5\), \(b=2\), and \(c=1\), so \(D=2^2-4(5)(1)=4-20=-16\). Therefore, option A is correct; since \(D<0\), the equation has no real roots. Exam tip: identify \(a\), \(b\), and \(c\) carefully, including their signs, before applying the formula.

Because (D=-16<0), the number of real roots is (0). In exams write this conclusion as soon as (D<0) appears.

Here, \(a=9\), \(b=-6\), and \(c=1\). The discriminant is \(D=b^2-4ac=(-6)^2-4(9)(1)=36-36=0\). When \(D=0\), a quadratic equation has two real and equal roots. In fact, \((3x-1)^2=0\), so both roots are \(x=\frac{1}{3}\). “Two positive and distinct” is incorrect because the root is positive but repeated, not distinct. 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, \(a=2, b=5, c=2\), 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 \(D=0\).