Extra Questions for Class 10 Maths Chapter 4 Quadratic Equations Based on Nature of Roots

Extra Questions for Class 10 Maths Chapter 4 Quadratic Equations

Nature of Roots of a Quadratic Equation

Let the given equation be ax² + bx + c = 0, where a = 0.

Then, the discriminant is given by

D = (b² – 4ac) . And, the roots of the given equation are

Case 1: If D > 0, then the roots are real and distinct.

These roots are given by,

Case 2: If D = 0, then In this case, the roots are real and equal.

Each root = -b/2a

Case 3: If D < 0, then the roots are imaginary. So, the given equation has no real roots.

Question: Show that the equation 2×2 – 6x + 3 = 0 has real roots.

Solution:

The given equation is 2x² – 6x + 3 = 0.
This is of the form ax² + bx + c = 0, where a = 2, b = -6 and c = 3.
Therefore, D = (b² – 4ac) = [(-6)² – 4 x 2 x 3] = (36 – 24) = 12 > 0.

So, the given equation has real unequal roots.

Question: Find the value of k for which the roots of the quadratic equation kx(x – 2) + 6 = 0 are equal.

Solution:

The given equation is kx² – 2kx + 6 = 0.
This is of the form ax² + bx + c = 0, where a = k, b = -2k and c = 6.
D = (b² – 4ac) = (4k² – 4 x k x 6) = (4k² – 24k).
For equal roots, we must have
D = 0

⇒ 4k² – 24k=0

⇒ 4k(k – 6) = 0

Either k = 0 or k = 6.
Now, k ≠ 0, we get 6 = 0, which is not possible.
Therefore, k = 6.

Question: Determine the value of k, (k > 0) such that both the equations x² + kx + 64 = 0 and
x² –8x + k = 0 will have real roots

Solution:

$$ x^2+k x+64=0 $$ To have real roots $$ \begin{aligned} b^2-4 a c \geq 0 \\ k^2-256 \geq 0 \\ \Rightarrow(k+16)(k-16) \geq 0 \quad & \\ \end{aligned} \\ \Rightarrow \quad k \geq 16 \text { or } k \geq-16\\ x^2-8 x+k=0 $$ To have real roots $$ \begin{aligned} & b^2-4 a c \geq 0 \\ & (-8)^2-4 \times 1 \times k \geq 0 \\ & 64-4 k \geq 0 \quad \Rightarrow \quad 4 k \leq 64 \quad \Rightarrow \quad k \leq 16 \end{aligned} $$

since k > 0, therefore 0 < k ≤ 16 for both equations to have real roots both will satisfy k = 16

Question: Determine the value of k so that the following linear equations has no solution.
(3x + 1) x + 3y – 2 = 0
(k² + 1) x + (k – 2) y – 5 = 0

Solution:

Here $$a_1=3 \mathrm{k}+1, \mathrm{~b}_1=3$$ and $$\mathrm{c}_1=-2$$ $$ \mathrm{a}_2=\mathrm{k}^2+1, \mathrm{~b}_2=\mathrm{k}-2 \text { and } \mathrm{c}_2=-5 $$ For no solution, condition is $$\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}$$ $$ \begin{aligned} & \frac{3 \mathrm{k}+1}{\mathrm{k}^2+1}=\frac{3}{\mathrm{k}-2} \neq \frac{-2}{-5} \\ & \Rightarrow \frac{3 \mathrm{k}+1}{\mathrm{k}^2+1}=\frac{3}{\mathrm{k}-2} \text { and } \frac{3}{\mathrm{k}-2} \neq \frac{2}{5} \end{aligned} $$ Now, $$\quad \frac{3 \mathrm{k}+1}{\mathrm{k}^2+1}=\frac{3}{\mathrm{k}-2}$$ $$ \begin{aligned} & \Rightarrow \quad(3 \mathrm{k}+1)(\mathrm{k}-2)=3\left(\mathrm{k}^2+1\right) \\ & \Rightarrow \quad 3 \mathrm{k}^2-5 \mathrm{k}-2=3 \mathrm{k}^2+3 \\ & \Rightarrow \quad-5 \mathrm{k}-2=3 \\ & \Rightarrow \quad-5 \mathrm{k}=5 \\ & \Rightarrow \mathrm{k}=-1 \end{aligned} $$ Clearly, $$\frac{3}{k-2} \neq \frac{2}{5} \text { for } k=-1 $$ Hence, the given system of equations will have no solution for k = – 1.

Q.1. Show that the roots of the equation x² + px – q² = 0 are real for all real values of p and q.

Q.2. Find the nature of the roots of the following quadratic equations:
(i) 2x² – 8x + 5 = 0 (iv) 5x(x – 2) + 6 = 0 (v) 12x² – 4√15 x + 5 = 0

Ans. (i) Real and unequal (iv) Not real (v) Real and equal

Q.3. For what value of k are the roots of the quadratic equation kx(x – 2√5) + 10 = 0 real and equal?

Ans. k = 0 or k = 2

Q.4. If the quadratic equation (1 + m²)x² + 2mcx + c² – a² = 0 has equal roots, prove that c² = a² (1 + m²) .

Q.5. (i) Find the values of k for which the quadratic equation (3k + 1)x² + 2(k + 1)x + 1 = 0 has real and equal roots.
(ii) Find the value of k for which the equation x² + k(2x + k – 1) + 2 = 0 has real and equal roots.

Ans. (i) k = 0 or k = 1 (ii) k = 2

Q.6. If –5 is a root of the quadratic equation 2x² + px – 15 = 0 and the quadratic equation p(x² + x) + k = 0 has equal roots, fi nd the value of k.

Ans. 49/28

Q.7. For what values of k are the roots of the quadratic equation 3x² + 2kx + 27 = 0 real and equal?

Ans. k = 9 or k = -9

Q.8. If the roots of the equations ax² + 2bx + c = 0 and bx² – 2√ac x + b = 0 are simultaneously real then prove that b² = ac.

Q.9. If –4 is a root of the equation x² + 2x + 4p = 0, fi nd the value of k for which the quadratic equation x² + px(1 + 3k) + 7(3 + 2k) = 0 has equal roots.

Ans. k=2 or k=-10/9

Q.10. Find the value of α for which the equation (α – 12)x² + 2(α – 12)x + 2 = 0 has equal roots.

Ans. α=14

Q.11. Find the values of k for which the given quadratic equation has real and distinct roots:
(i) kx² + 6x + 1 = 0
(ii) x² – kx + 9 = 0
(iii) 9x² + 3kx + 4 = 0

Ans. (i) k < 9 (ii) k > 6 or k < –6 (iii) k > 4 or k < –4

Q.12. For what values of p are the roots of the equation 4×2 + px + 3 = 0 real and equal?

Ans. p = 4√3 or p = -4√3

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