# AP SSC 10th Class Maths Solutions Chapter 5 Quadratic Equations InText Questions

## AP State Syllabus SSC 10th Class Maths Solutions 5th Lesson Quadratic Equations InText Questions

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 5 Quadratic Equations InText Questions and Answers.

### 10th Class Maths 5th Lesson Quadratic Equations InText Questions and Answers

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Question 1.
Check whether the following equations are quadratic or not. (Page No. 102)
i) x2 – 6x – 4 = 0
ii) x3 – 6x2 + 2x – 1 = 0
iii) 7x = 2x2
iv) x2 + $$\frac{1}{\mathbf{x}^{2}}$$ = 2
v) (2x + 1) (3x + 1) = 6(x – 1) (x – 2)
vi) 3y2 = 192
Answer:
i) x2 – 6x – 4 = 0
Yes. It’s a quadratic equation.
ii) x3 – 6x2 + 2x – 1 =0
No. It is not a quadratic equation. [∵ degree is 3]
iii) 7x = 2x2
Yes. It’s a quadratic equation.
iv) 2 + $$\frac{1}{\mathbf{x}^{2}}$$ = 2
No. It is not a quadratic equation. [∵ degree is 4]
v) (2x + 1) (3x + 1) = 6(x – 1) (x – 2) No. It is not a quadratic equation, [∵ co-efficient of x2 on both sides is same i.e. 6]
vi) 3y2 = 192
Yes. It’s a quadratic equation.

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Question 1.
Verify that 1 and $$\frac{3}{2}$$ are the roots of the equation 2x2 – 5x + 3 = 0. (Page No. 107)
Answer:
Let the given Q.E. be p(x) = 2x2 – 5x + 3
Now p(1) = 2(1)2 – 5(1) + 3
= 2 – 5 + 3 = 0
∴ 1 is a root of 2x2 – 5x + 3 = 0
also p$$\left(\frac{3}{2}\right)$$ = 2$$\left(\frac{3}{2}\right)^{2}$$ – 5$$\left(\frac{3}{2}\right)$$ + 3
= 2 × $$\frac{9}{4}$$ – $$\frac{15}{2}$$ + 3
= $$\frac{9}{2}$$ + 3 – $$\frac{15}{2}$$
= $$\frac{9+6-15}{2}$$ = 0
∴ $$\frac{3}{2}$$ is also a root of 2x2 – 5x + 3 = 0.

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Question 1.
Solve the equations by completing the square.  (Page No. 113)
i) x2 – 10x + 9 = 0
Answer:
Given: x2 – 10x + 9 = 0
⇒ x2 – 10x = -9
⇒ x2 – 2.x.5 = -9
⇒ x2 – 2.x.5 + 52 = -9 + 52
⇒ (x – 5)2 = 16
∴ x – 5 = ± 4
∴ x – 5 = 4 (or) x – 5 = -4
⇒ x = 9 (or) x = 1
⇒ x = 9 (or) 1 ii) x2 – 5x + 5 = 0
Answer:
Given: x2 – 5x + 5 = 0
⇒ x2 – 5x = 5 iii) x2 + 7x-6 = 0
Answer:
x2 + 7x – 6 = 0
x2 + 7x = 6   Think & Discuss

Question 1.
We have three methods to solve a quadratic equation. Among these three, which method would you like to use 7 Why? (Page No. 115)
Answer:
If the Q.E. has distinct and real roots, we use factorisation. If Q.E. has no real roots, we use quadratic formula.

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Question 1.
Explain the benefits of evaluating the discriminant of a quadratic equation before attempting to solve it. What does its value signifies?  (Page No. 122)
Answer:
The discriminant b2 – 4ac of a Q.E.
ax2 + bx + c = 0 gives the clear idea about the nature of the roots of the Q.E.
If the discriminant D = b2 – 4ac > 0, the Q.E. has distinct and real roots.
If b2 – 4ac = 0, the Q.E. has equal roots.
If b2 – 4ac < 0, the Q.E. has no real roots.
By the value of the discriminant, we can state the nature of the roots of a Q.E. without actually finding them. Question 2.
Write three quadratic equations one having two distinct real solutions, one having no real solution and one having exactly one real solution.  (Page No. 122)
Answer: 