Nature of Roots

Here, \(a=1\), \(b=14\), and \(c=49\). Thus, the discriminant is \(D=b^2-4ac=14^2-4(1)(49)=196-196=0\). When \(D=0\), the two roots are real and equal. In fact, the equation can be written as \((x+7)^2=0\), giving the repeated root \(-7\). Exam tip: choose equal real roots for \(D=0\); distinct real roots require \(D>0\).

Here, \(a=1\), \(b=-4\), and \(c=8\). The discriminant is \(D=b^2-4ac=(-4)^2-4(1)(8)=16-32=-16\). Since \(D<0\), the equation has no real roots. In fact, its roots are \(2+2i\) and \(2-2i\), which are non-real and distinct. Exam tip: For a quadratic equation, \(D<0\) means that there are no real roots.

The governing condition for equal real roots of ax²+bx+c=0 is that the discriminant must be zero: b²−4ac=0. Comparing x²−3x+s=0 with the standard form gives a=1, b=−3, and c=s. Therefore, (−3)²−4(1)(s)=0, so 9−4s=0 and s=9/4. The repeated root can independently confirm the result. Since the root is −b/(2a)=3/2, substitution gives (3/2)²−3(3/2)+9/4=9/4−9/2+9/4=0. Thus option A is correct. The values 3/4, 9, and 3 make the discriminant nonzero, so they produce two unequal roots rather than a repeated real root.

For a quadratic equation \(ax^2+bx+c=0\) to have real and equal roots, its discriminant must be zero: \(D=b^2-4ac=0\). Here, \(a=5\), \(b=-4\), and \(c=t\), so \((-4)^2-4(5)t=16-20t=0\). Therefore, \(t=\frac{4}{5}\). Option B results from reversing the numerator and denominator. Exam tip: whenever roots are real and equal, immediately use \(D=0\).

For \(x^2+1=0\), \(a=1, b=0, c=1\). Thus, the discriminant is \(D=b^2-4ac=0^2-4(1)(1)=-4<0\). A quadratic equation with a negative discriminant has no real roots, so the statement is correct. Exam tip: For \(D<0\), write ‘no real roots.’

For a quadratic equation, the discriminant is \\(D=b^2-4ac\\). Here, \\(a=3, b=10, c=3\\), so \\(D=10^2-4(3)(3)=64\\). Since \\(D>0\\), the two roots are real and distinct. If \\(D=0\\), the roots would be real and equal, so option B is incorrect. Exam tip: The sign of the discriminant directly determines the nature of the roots.

The nature of the roots of a quadratic equation is determined by its discriminant, \(D=b^2-4ac\). If \(D>0\), the roots are real and distinct; if \(D=0\), the roots are real and equal; and if \(D<0\), there are no real roots. Therefore, the discriminant is the correct quantity. Exam tip: first write the equation in standard form and identify \(a\), \(b\), and \(c\) before calculating \(D\).

For a quadratic equation, the discriminant is \(D=b^2-4ac\). When \(D>0\), \(\sqrt{D}\) is positive and real, so the two signs in \(x=\frac{-b\pm\sqrt{D}}{2a}\) give two distinct real roots. Equal roots occur when \(D=0\), while no real roots occur when \(D<0\); therefore, options B and C are incorrect. In an exam, first check the sign of the discriminant to determine the nature of the roots.

The discriminant is \(D=b^2-4ac\). When \(D<0\), the root formula contains \(\sqrt{D}\), the square root of a negative number. Hence, there are no real roots; the roots form a complex conjugate pair. Exam tip: \(D>0\) gives two distinct real roots.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=1\), \(b=-5\), and \(c=6\), so \(D=(-5)^2-4(1)(6)=25-24=1\). The value 0 would indicate equal roots, but this equation has \(D=1\). Exam tip: identify the signs of \(a\), \(b\), and \(c\) carefully before substituting.

Here, \(a=1, b=-4, c=4\). The discriminant is \(D=b^2-4ac=(-4)^2-4(1)(4)=0\), so the roots are real and equal. In fact, \(x^2-4x+4=(x-2)^2\), giving the repeated root \(x=2\). Exam tip: when \(D=0\), the roots are always real and equal.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=2\), \(b=-3\), and \(c=1\), so \(D=(-3)^2-4(2)(1)=9-8=1\). Therefore, the correct answer is 1. In exams, be careful to square the negative value of \(b\) correctly.

The governing criterion for the nature of roots of ax²+bx+c=0 is the discriminant D=b²−4ac. In 2x²−3x+1=0, a=2, b=−3 and c=1. Therefore D=(−3)²−4(2)(1)=9−8=1. Since D is positive, the equation has two real and distinct roots. Direct factorisation confirms this: 2x²−3x+1=(2x−1)(x−1), so the roots are x=1/2 and x=1. They are both real and unequal. Equal real roots require D=0, whereas non-real roots require D<0. Hence option A is the only correct choice. The factorisation also provides an independent check that the discriminant-based conclusion is correct.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=3, b=6, c=3\), so \(D=6^2-4(3)(3)=0\). When \(D=0\), the roots are real and equal; in fact, both roots are \(-1\). Therefore, option A is correct. Exam tip: \(D=0\) always indicates equal real roots.

For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Here, \(a=1,b=1,c=1\), so \(D=1^2-4(1)(1)=-3\), which is negative. Therefore, option A is correct. The discriminant would be zero only if \(D=0\), which is not the case here. Exam tip: when \(D<0\), the quadratic equation has no real 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 a quadratic ax²+bx+c=0, the discriminant D=b²−4ac determines the nature of its roots. In x²−2x−3=0, the coefficients are a=1, b=−2, and c=−3. Thus D=(−2)²−4(1)(−3)=4+12=16, which is positive. A positive discriminant means two real and distinct roots. The result can also be checked by factorisation: x²−2x−3=(x−3)(x+1), so the roots are 3 and −1. They are real and unequal, and one is positive. Equal roots require D=0, while non-real roots require D<0. Therefore option A is correct and option D is contradicted by the actual roots.

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.