Here you get the CBSE Class 10 Mathematics chapter 4, Quadratic Equations: NCERT Exemplar Problems and Solutions (Part-II). This part of the chapter includes solutions for Exercise 4.2 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Quadratic Equations. This exercise comprises of only the Short Answer Type Questions framed from various important topics in the chapter. Each question is provided with a detailed solution.

NCERT Exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. All these questions are very important to prepare for CBSE Class 10 Mathematics Board Examination 2017-2018 as well as other competitive exams.

**Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Quadratic Equations:**

**Exercise 4.2**

**Very Short Answer Type Questions**

**Question11.** State whether the following quadratic equations have two distinct real roots. Justify your answer.

**Solution**:

A quadratic equation, *ax*^{2} + *bx* + *c* = 0; *a* ≠ 0 will have two distinct real roots if its discriminant, *D* = *b*^{2} - 4*ac* > 0.

**(i)** Given equation is *x*^{2} – 3*x* + 4 = 0

On comparing with standard equation we have,

*a* = 1, *b* = –3 and *c* = 4

Now, *D* = *b*^{2} – 4*ac* = (–3)^{2} –4(1) (4) = 9 – 16 = – 7< 0

Hence, the equation *x*^{2} –3*x* + 4 = 0 has no real roots.

**(ii)** Given equation is, 2*x*^{2} + *x* – 1 = 0

On comparing with standard equation we have,

*a* = 2, *b* = 1 and *c* = –1

Now, *D** *= *b*^{2} – 4*ac* = (1)^{2} – 4(2) (–1) = 1 + 8 = 9 > 0

Hence, the equation 2*x*^{2} + *x* – 1 = 0 has two distinct real roots.

**(iv)** Given, 3*x*^{2} –4*x* + 1 = 0

On comparing with standard equation we have,

*a* = 3, *b* = –4 and *c* = 1

Now, *D** *= *b*^{2} –4*ac* = (–4)^{2} – 4(3) (1) = 16 – 12 = 4 > 0

Hence, the equation 3*x*^{2} – 4*x* + 1 = 0 has two distinct real roots.

**(v)** Given, (*x* + 4)^{2} –8*x* = 0

Simplifying the above equation, we have:

*x*^{2}+ 16 + 8*x* – 8*x* = 0 [∵(*a* + *b*)^{2} = *a*^{2}+ 2*ab* + *b*^{2}]

⟹ *x*^{2} + 16 = 0

⟹ *x*^{2} + 0. *x* + 16 = 0

On comparing with standard equation we have:

*a* = 1, *b* = 0 and *c* = 16

Now, *D** *= *b*^{2} –4*ac* = (0)^{2} – 4 (1) (16) = – 64 < 0

Hence, the equation (*x* + 4)^{2} – 8*x* = 0 has imaginary roots.

**(viii)** Given, *x* (1 - *x*) -2 = 0

Simplifying above equation we have:

⟹ *x* - *x*^{2} − 2 = 0

⟹ *x*^{2} – *x* + 2 = 0

On comparing with standard equation we have,

*a* = 1, *b* = –1 and *c* = 2

Now, *D** *= *b*^{2} – 4*ac* = (-1)^{2} –4(1) (2)=1 – 8 = – 7 < 0

Hence, the equation *x* (1 – *x*) –2 = 0 has imaginary roots.

**(ix)** Given, (*x *– 1) (*x* + 2) + 2 = 0

Simplifying above equation we have:

⟹ *x*^{2} + *x* – 2 + 2 = 0

⟹ *x*^{2} + *x* + 0 = 0

On comparing with standard equation we have,

*a* = 1, *b* = 1 and *c* = 0

Now, *D** *= *b*^{2} – 4*ac* = 1 – 4(1) (0) = 1 > 0

Hence, equation (*x *– 1) (*x* + 2) + 2 = 0 has two distinct real roots.

**(x)** Given, (*x* + 1) (*x* – 2) + *x* = 0

Simplifying above equation we have:

⟹ *x*^{2} + *x* – 2*x* – 2 + *x* = 0

⟹ *x*^{2} – 2 = 0

⟹ *x*^{2} + 0 *x* – 2 = 0

On comparing with *ax*^{2} + *bx* + *c* = 0, we have

a = 1, *b* = 0 and *c* = –2

Now, *D** *= *b*^{2} –4*ac* = (0)^{2} – 4(1) (–2) = 0 + 8 = 8 > 0

Hence, the equation (*x* + 1) (*x* – 2) + *x* = 0 has two distinct real roots.

**Question 2.** Write whether the following statements are true or false. Justify your answers.

(i) Every quadratic equation has exactly one root.

(ii) Every quadratic equation has atleast one real root.

(iii) Every quadratic equation has atleast two roots.

(iv) Every quadratic equation has almost two roots.

(v) If the coefficient of *x*^{2} and the constant term of a quadratic equation have opposite signs, then the quadratic equation has real roots.

(vi) If the coefficient of *x*^{2}and the constant term have the same sign and if the coefficient of *x* term is zero, then the quadratic equation has no real roots.

**Solution**:

(i) False. For example, a quadratic equation *x*^{2} – 9 = 0 has two distinct roots – 3 and 3.

(ii) False. For example, equation *x*^{2} + 4 = 0 has no real root.

(iii) False. For example, a quadratic equation *x*^{2} – 4*x *+ 4 = 0 has only one root which is 2.

(iv) True, because every quadratic polynomial has almost two roots.

(v) True, because in this case discriminant is always positive. For example, in *ax*^{2}+ *bx* + *c* = 0, as *a* and *c* have opposite sign so, *ac* < 0 ⟹ Discriminant = *b*^{2} – 4*ac* > 0.

(vi) True, because in this case discriminant is always negative. For example, in *ax*^{2}+ *bx* + *c* = 0, as *b* = 0, and *a* and *c* have same sign then *ac* > 0 ⟹ discriminant = *b*^{2} – 4*ac* = – 4 *a c* < 0

**Question3.** A quadratic equation with integral coefficient has integral roots. Justify your answer.

**Solution**:

No, a quadratic equcation with integral coefficient (0, ± 1, ± 2, ± 3…) can have fractional or non integral roots.

For example, the quadratic equation 8*x*^{2} – 2*x* – 1 = 0 have integral coefficients (8, −2 and −1)

**Question4.** Does there exist a quadratic equation whose coefficients are rational but both of its roots are irrational? Justify your answer.

**Solution**:

Yes, a quadratic equation having coefficients as rational number, has irrational roots.

For example, the quadratic equation *x*^{2} – 4*x* – 3 = 0 has rational coefficients.

Now, *b*^{2} – 4*ac* = 16 + 12 = 28.

**Question5.** Does there exist a quadratic equation whose coefficient are all distinct irrationals but both the roots are rationals? Why?

**Solution**:

Yes, there may be a quadratic equation whose coefficients are all distinct irrationals, but both the roots are rational.

Hence roots are rational.

**Question6.** Is 0.2 a root of the equation *x*^{2} –0.4 = 0? Justify your answer.

**Solution**:

No, because 0.2 does not satisfy the given quadratic equation, as,

(0.2)^{2} – 0.4 = 0.04 – 0.4 = − 0.36 ≠ 0.

**Question7.** If *b* = 0, *c* < 0, is it true that the roots of *x*^{2} + *bx* + *c* = 0 are numerically equal and opposite in sign? Justify your answer.

**Solution**:

Given quadratic equation is, *x*^{2} + *bx* + *c* = 0 ...(i)

Also given that *b* = 0 and *c* < 0

Putting *b* = 0 in equation (i), we get:

Hence, roots of *x*^{2} + *bx* +*c* = 0 are numerically equal and opposite in sign.

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