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CBSE Class 10 Mathematics — practice by topic

Browse 710 questions across 9 Mathematics topics. Choose what you want to revise, then open a question to check its answer and worked solution.

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Mathematics topics

Introduction to Trigonometry (131)Mensuration: Surface Areas & Volumes (104)Real Numbers and Polynomials (100)Arithmetic Progression (98)Statistics and Probability (77)Coordinate Geometry (50)Linear Equations in Two Variables (50)Quadratic Equations (50)Triangles and Circles (50)

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If the discriminant of a quadratic equation is negative, what can be concluded about its roots?Read the answer and worked solution →For the quadratic equation $ax^2+bx+c=0$, what is the discriminant?Read the answer and worked solution →Find the roots of $2x^2-5x+3=0$ by factorisation.Read the answer and worked solution →What is the maximum number of real roots a quadratic equation can have?Read the answer and worked solution →

Worked practice questions

If the discriminant of a quadratic equation is negative, what can be concluded about its roots?

  1. The equation has two distinct real roots
  2. The equation has two equal real roots
  3. The equation has no real roots
  4. The equation has exactly one real root
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Correct answer: C. The equation has no real roots

If $b^2-4ac<0$, then the square root term in the quadratic formula is not a real number. **Therefore, the quadratic equation has no real roots.**

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For the quadratic equation $ax^2+bx+c=0$, what is the discriminant?

  1. $b^2+4ac$
  2. $4ac-b^2$
  3. $b^2-4ac$
  4. $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
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Correct answer: C. $b^2-4ac$

The discriminant of the quadratic equation $ax^2+bx+c=0$ is the expression under the square root in the quadratic formula, namely $b^2-4ac$. It determines the nature of the roots.

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Find the roots of $2x^2-5x+3=0$ by factorisation.

  1. $x=\frac{3}{2},\ 1$
  2. $x=\frac{5}{2},\ 3$
  3. $x=-\frac{3}{2},\ -1$
  4. $x=\frac{2}{3},\ 1$
Show answer and solution

Correct answer: A. $x=\frac{3}{2},\ 1$

Factorise $2x^2-5x+3$ by splitting the middle term: $2x^2-2x-3x+3=2x(x-1)-3(x-1)=(2x-3)(x-1)$. So $(2x-3)(x-1)=0$, giving $2x-3=0$ or $x-1=0$. **Hence the roots are $x=\frac{3}{2}$ and $x=1$.**

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