Q.1 Solve the following quadratic equations by factorisation

(i) x²+6x+5=0

(ii) 9x²-3x-2=0

(iii) x²-8x+16=0

(iii) \frac{1}{x-2} +\frac{2}{x-1} =\frac{6}{x}

(v) x²+2√2x-6=0

(vi) x²-2ax+a²-b²=0

(vii) \frac{1}{a+b+x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x}

Solution:

(i) We have ,

x²+6x+5=0

x²+5x+x+5=0

x(x+5)+(x+5)=0

(x+5)(x+1)=0

So,x=-5 or -1

(ii) We have ,

9x²-3x-2=0

9x²-6x+3x-2=0

3x(3x-2)+(3x-2)=0

(3x-2)(3x+1)=0

S0, x=\frac{2}{3} or \frac{-1}{3}

(iii) We have ,

x²-8x+16=0

x²-4x-4x+16=0

x(x-4)-4(x-4)=0

(x-4)(x-4)=0

So,x=4

(iv) We have ,

\frac{1}{x-2} +\frac{2}{x-1} =\frac{6}{x}

Take LCM ,

\frac{(x-1)+2(x-2)}{(x-2)(x-1)}=\frac{6}{x}

\frac{3x-5}{x²-3x+2}=\frac{6}{x}

3x²-13x+12=0

3x²-9x-4x+12=0

3x(x-3)-4(x-3)=0

(x-3)(3x-4)=0

So, x= 3 or \frac{4}{3}

(v) We have ,

x²+2√2x-6=0

Spiltting the middle term method

x²+3√2x-√2x-6=0

x(x+3√2)-√2(x+3√2)=0

(x+3√2)(x-√2)=0

So,x =-3√2 or √2

(vi) We have ,

x²-2ax+a²-b²=0

Factor of constant term=(a²-b²) = (a-b)(a+b)

Express middle term coefficient in the term of above factor =-2a=-{(a-b)+(a+b)}

x²-2ax+a²-b²=0

x²-{(a-b)+(a+b)}x+(a-b)(a+b)=0

x²-(a-b)x-(a+b)x+(a-b)(a+b)=0

x{x-(a-b)}-(a+b){x-(a-b)}=0

{x-(a-b)}{x-(a+b)}=0

So, x=(a-b) or (a-b)

(vii) We have ,

\frac{1}{a+b+x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x}

\frac{1}{a+b+x}\frac{1}{x}=\frac{1}{a}+\frac{1}{b}

\frac{x-(a+b+x)}{x(a+b+x)}=\frac{a+b}{ab}

⇒-ab(a+b)=(a+b)x(a+b+x)

⇒x(a+b+x)+ab=0

⇒x²+ax+bx+ab=0

⇒x(x+a)+b(x+a)=0

⇒(x+a)(x+b)=0

So,x=-a or -b

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Q.2 Find the values of k for which the following equation has equal roots

(k-12)x²+2(k-12)x+2=0

Solution: We have ,

(k-12)x²+2(k-12)x+2=0  where a=(k-12) ,b=2(k-12) and c=2

We know, Discriminant (D)=b²-4ac

Discriminant (D)={2(k-12)}²-4(k-12)(2)

Discriminant (D)=4{k²+144-24x}-8(k-12)

Discriminant (D)=4(k-12)(k-14)

SInce ,it is given that equation has equal roots .So,Discriminant (D)=0

4(k-12)(k-14) =0

(k-12)=0  or (k-14)=0

k=12   or k=14

Q.3 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 ,find the value of k

Solution: It is given that -5 is the root of quadratic equation f(x)=2x²+px-15=0

f(-5)=0

2(-5)²+p(-5)-15=0

2(25)-5p-15=0

50-5p-15=0

35-5p=0

p=7

Q.4 Find the values of k for which the given equation has real roots

(i) kx²-6x-2=0

(ii) 5x²-kx+1=0

Solution:

(i) We have kx²-6x-2=0 as a quadratic equation in which a =k ,b=-6 and c=-2

Discriminant(D)=b²-4ac=(-6)²-4(k)(-2)

Discriminant(D)=b²-4ac=36+8k

Since ,it has real roots so,Discriminant(D)≥0

36+8k≥0

k≥-\frac{36}{8}

(ii) We have 5x²-kx+1=0  as a quadratic equation in which a =5 ,b=-k and c=1

Discriminant(D)=b²-4ac=(-k)²-4(5)(1)

Discriminant(D)=b²-4ac=k²-20

Since ,it has real roots so,Discriminant(D)≥0

k²-20≥0

k≤-4 and k≥4