Nature of Roots

Comparing the equation with the standard form \\(ax^2+bx+c=0\\), we get \\(a=7\\), \\(b=2\\), and \\(c=3\\). Thus, \\(D=b^2-4ac=(2)^2-4(7)(3)=4-84=-80\\). Therefore, -80 is correct. Since \\(D<0\\), the equation has no real roots. In exams, remember to use the coefficient of \\(x\\), including its sign, when calculating \\(b^2\\).

Here, \(a=7\), \(b=2\), and \(c=3\). The discriminant is \(\Delta=b^2-4ac=2^2-4(7)(3)=4-84=-80\). Since \(\Delta<0\), the equation has no real roots. Two real and equal roots occur only when \(\Delta=0\), so option B is incorrect. In an exam, first check the sign of the discriminant to determine the nature of the roots.

For equal real roots, the discriminant must be \(D=0\). Here, \(a=1, b=-2r, c=9\), so \(D=b^2-4ac=(-2r)^2-4(1)(9)=4r^2-36\). Thus, \(4r^2-36=0\), giving \(r^2=9\). Exam tip: For equal roots, immediately use the condition \(D=0\).

For real and equal roots, the discriminant \(D=b^2-4ac\) must be zero. Here, \(a=4\), \(b=s\), and \(c=9\), so \(s^2-4(4)(9)=0\). Therefore, \(s^2=144\). Exam tip: For equal roots of a quadratic equation, set the discriminant equal to zero.

Here (D=(-5)^2-4(1)(3)=13), and (13) is not a perfect square. So the roots are real, irrational and distinct.

A quadratic equation \(ax^2+bx+c=0\) has no real roots when its discriminant \(D=b^2-4ac\) is negative. Here, \(a=1, b=t, c=12\), so \(D=t^2-48\). Thus, \(t^2-48<0\), which gives \(t^2<48\). If \(t^2=48\), the equation has two equal real roots, so option B is incorrect. Exam tip: determine the nature of the roots by checking the sign of the discriminant first.

The relevant concept is the discriminant of a quadratic equation. For ax²+bx+c=0, D=b²−4ac. Here a=1, b=−2, and c=−3, so D=(−2)²−4(1)(−3)=4+12=16. Since D>0, there are two real and distinct roots. Using the quadratic formula gives x=[2±√16]/2=(2±4)/2, hence x=3 and x=−1. Both roots are also rational, but the question asks primarily about their nature, so the complete classification is two real and unequal roots. Option B would require D=0, option C would require D<0, and option D is false because there are two rational roots rather than one. Therefore A is correct.

The standard form is ax²+bx+c=0, so here a=4, b=−4 and c=1. The discriminant is D=b²−4ac=(−4)²−4(4)(1)=16−16=0. A zero discriminant means that the two roots are real and equal. The equation also factors as 4x²−4x+1=(2x−1)², so 2x−1=0 and the repeated root is x=1/2. Therefore option A gives both the correct nature and the correct value. Option B incorrectly states D=4 and distinct roots; option C contradicts D=0 because real repeated roots exist; and option D uses an incorrect discriminant and incorrectly calls the root irrational. The factorisation independently confirms the discriminant result.

For 2x²+3x−5=0, identify a=2, b=3 and c=−5. The discriminant is D=b²−4ac=3²−4(2)(−5)=9+40=49. Since D is positive, the roots are real and distinct. Since 49 is a perfect square, the square root in the quadratic formula is rational, so both roots are rational. Directly, x=[−3±√49]/(2·2)=[−3±7]/4, which gives x=1 and x=−5/2. Thus option A is fully correct. Option B would require D=0, option C would require a negative discriminant, and option D is wrong because D=7 is not the calculated value and the roots are rational rather than irrational.

For a quadratic equation Ax² + Bx + C = 0, the discriminant D = B² − 4AC determines the nature of its roots. Here A = 1, B = −9, and C = 20. Thus D = (−9)² − 4(1)(20) = 81 − 80 = 1. Since D is positive, the roots are real and distinct. Moreover, D = 1 is a perfect square, and with rational coefficients this means both roots are rational. Indeed, the equation factors as (x − 4)(x − 5) = 0, giving roots 4 and 5. Therefore option A is correct; D = 0 would indicate equal roots, while D < 0 would indicate no real roots.

For a quadratic equation ax²+bx+c=0, the discriminant is D=b²−4ac. Comparing 5x²−2x+3=0 with the standard form gives a=5, b=−2, and c=3. Substitution yields D=(−2)²−4(5)(3)=4−60=−56. Since −56 is less than zero, the discriminant is negative, so option C is correct. This sign also indicates that the equation has no real roots and two non-real complex conjugate roots. Option A would require the calculated value to be positive, while option B would require it to be exactly zero. Option D is simply not the result of the discriminant formula. The requested conclusion follows directly from the sign of −56.