Question

ʜᴇʟʟᴏ

ᴍᴀᴛʜꜱ
ᴄʟᴀꜱꜱ 10
ᴄʜ - qᴜᴀᴅʀᴀᴛɪᴄ ᴇqᴜᴀᴛɪᴏɴꜱ
•ᴩʟᴇᴀꜱᴇ ᴀɴꜱᴡᴇʀ ᴛʜᴇ qᴜᴇꜱᴛɪᴏɴ.
•ᴩʟᴇᴀꜱᴇ ᴀɴꜱᴡᴇʀ ᴀʟʟ ᴛʜᴇ ᴩᴀʀᴛꜱ ᴏꜰ qᴜᴇꜱᴛɪᴏɴ.

ᴛʜᴀɴᴋꜱ​

Answer 1

QUESTION :

Solve the following quadratic equations by factorisation method :

SOLUTION :

1. $4x^{2} - 4a^{2}x + (a^{4} - b^{4}) = 0$

→ $4x^{2} - 4a^{2}x + a^{4} - b^{4} = 0$

Using $\sf (a-b)^{2} = a^{2} - 2ab + b^{2}$

$(2x - a^{2}) - b^{4}$

Using $(a-b)(a+b) = a^{2} - b^{2}$

→ $(2x - a^2 - b^2) \times (2x - a^{2} + b^{2}$

→ $x = \dfrac {a^{2} - b^{2}}{2} or x = \dfrac {a^{2} + b^{2}}{2}$

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2. $9x^{2} - 9 (a+b)x + (2a^{2} + 5ab + 2b^{2})$

Multiplying brackets,

→ $9x^{2} - 9ax + 9bx + (2a^{2} + 5ab + 2b^{2})$

Remove the brackets,

→ $9x^{2} - 9ax - 9bx + 2a^{2} + 5ab + 2b^{2}$

→ $x = 2a + \dfrac {b}{3} or x = 2a + \dfrac {2b}{3}$

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3. $a^{2}b^{2}x^{2} + b^{2}x - a^{2}x - a^{2}x - 1$

→ $b^{2}x \times (a^{2}x + 1) - a^{2}x - 1$

→ $b^{2}x \times (a^{2}x + 1) - (a^{2}x + 1)$

→ $(a^{2}x + 1) \times (b^{2}x - 1)$

→ $x = \dfrac {1}{b^{2}} and x = \dfrac {-1}{a^{2}}$

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4. $\dfrac {3}{x+1} + \dfrac {3}{2}(x+1) = \dfrac {29}{4x-1}$

→ $3(x-1) + \dfrac {4(x+1)}{x^{2} - 1} = \dfrac {29}{4x-1}$

→ $\dfrac {7x+1}{x^{2} - 1} = \dfrac {29}{4x-1}$

→ $-x^{2} - 3x + 28 = 0$

→ $(x+7) (x-4) = 0 </p><p>→ [tex] x = -7 \: and \: x = 4$

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5. $\dfrac {a}{x-b} - \dfrac {b}{x-a} = 2 where x \neq a & b$

→ $2x^{2} - 3x(a+b) + (a+b)^{2} + (a+b)^{2} = 0$

→ $2x = a + b or x = a + b$

→ $x = \dfrac {a+b}{2} \: or \: x = a + b$

Answer 2

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