# 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
i) x2 – 6x – 4 = 0
ii) x3 – 6x2 + 2x – 1 =0
No. It is not a quadratic equation. [∵ degree is 3]
iii) 7x = 2x2
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

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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)
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
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
Given: x2 – 5x + 5 = 0
⇒ x2 – 5x = 5 iii) x2 + 7x-6 = 0
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)
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)  