What will be the nature of roots of (7x^2-5x-3=0)?
Here (D=(-5)^2-4(7)(-3)=109). (109) is positive but not a perfect square, so the roots are irrational and distinct.
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SubjectsMathematics
मूलों की प्रकृति
In Class 10 Mathematics, Nature of Roots explains how to determine the type and number of solutions of a quadratic equation. Students use the discriminant, b² − 4ac, to identify whether an equation has two distinct real roots, two equal real roots, or no real roots. The topic connects algebraic calculations with the graph of a quadratic function and helps learners interpret equations, compare cases, and solve related problems from the chapter Quadratic Equations.
TOPIC PRACTICE
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Here (D=(-5)^2-4(7)(-3)=109). (109) is positive but not a perfect square, so the roots are irrational and distinct.
View question detailsHere, \(a=3\), \(b=-5\), and \(c=2\). The discriminant is \(\Delta=b^2-4ac=(-5)^2-4(3)(2)=25-24=1\). Since \(\Delta>0\), the equation has two distinct real roots. Two equal real roots occur only when \(\Delta=0\), so option B is incorrect. Exam tip: calculate the discriminant first to determine the nature of the roots.
View question detailsHere (D=6^2-4\cdot1\cdot9=0), so the roots are equal real roots. In exams, (D=0) directly indicates equal roots.
View question detailsFor a quadratic equation ax²+bx+c=0, the discriminant D=b²−4ac determines the nature of its roots. Here a=2, b=3 and c=5, so D=3²−4(2)(5)=9−40=−31. Because D is negative, the equation has no real roots; its two roots are complex conjugates. Therefore option C is correct. Equal real roots would require D=0, while two distinct real roots would require D>0. Option D is also incorrect because a non-degenerate quadratic equation does not have exactly one real root under the usual discriminant classification; the negative discriminant gives no real root at all.
View question detailsFor equal roots, (D=0), so (4^2-4\cdot k\cdot1=0) gives (k=4). In exams, use (D=0) when equal roots are given.
View question detailsFor equal roots, (D=0), so ((-2k)^2-4\cdot1\cdot9=0) gives (k=\pm3). In exams, when (k^2) appears, check both signs.
View question detailsFor two distinct real roots, the discriminant (D=b^2-4ac) is positive. In exams, first calculate (D) to decide the nature.
View question detailsWhen (D=0), both roots are equal and real. In exams, these may also be called repeated roots.
View question detailsFor ax²+bx+c=0, the discriminant is D=b²−4ac. In this equation, a=5, b=−2 and c=3. Substitution gives D=(−2)²−4(5)(3)=4−60=−56. Since −56 is less than zero, the sign of D is negative, so option C is correct. The question asks only for the sign, not for the roots themselves; nevertheless, a negative discriminant also tells us that the equation has no real roots. Option A would correspond to a positive value, option B to exactly zero, and option D is not the result of the discriminant calculation.
View question detailsFor equal roots, (D=0), so (m^2-64=0) and (m=\pm8). In exams, remember (m^2=a^2) gives (m=\pm a).
View question detailsFor two distinct real roots, (D>0), so (25-4k>0) gives (k<\frac{25}{4}). In exams, keep the inequality sign correct while solving.
View question detailsFor no real roots, (D<0), so (k^2-64<0), that is (k^2<64). In exams, connect (D<0) with no real roots.
View question detailsHere (D=2^2-4\cdot7\cdot(-3)=88), so (D>0). In exams, when (c) is negative, (D) can often become positive.
View question detailsHere (D=12^2-4\cdot9\cdot4=0), so the roots are real and equal. In exams, (9x^2+12x+4) can also be recognized as ((3x+2)^2).
View question detailsA quadratic equation has two equal real roots precisely when its discriminant is zero, provided the coefficient of x² is nonzero. For px²+6x+3=0, a=p, b=6 and c=3. Thus D=b²−4ac=6²−4(p)(3)=36−12p. Set D=0: 36−12p=0, so 12p=36 and p=3. This value also satisfies p≠0, so the equation remains genuinely quadratic. Therefore option C is correct. The other proposed values give D=24, 12 and −12 respectively, so they produce distinct real roots for p=1 or 2 and no real roots for p=4, not equal roots.
View question detailsHere (D=(-7)^2-4\cdot1\cdot10=9), which is a positive perfect square. In exams, positive perfect-square (D) gives rational and distinct roots.
View question detailsHere (D=(-4)^2-4\cdot1\cdot1=12), which is positive but not a perfect square. In exams, such (D) gives irrational distinct real roots.
View question details(D=18) is positive but not a perfect square, so the roots are real, distinct, and irrational. In exams, also check whether (D) is a perfect square.
View question detailsFor equal roots, (D=0), so (q^2-144=0) and (q=\pm12). In exams, both signs of (b) may be possible.
View question detailsHere (D=(-11)^2-4\cdot6\cdot3=49), a positive perfect square. In exams, roots are rational in such cases.
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