SOLVING QUADRATIC EQUATION BY FACTORING PRACTICE

Solve the following quadratic equations by factorization method.

(i)  (2x + 3)²-81  =  0  

(ii)  3x²–5x–12  =  0  

(iii) √5 x²+2x–3√5  =  0 

(iv) 3(x²–6)  =  x(x+7)–3

(v) 3x–(8/x)  =  2

(vi) x+(1/x)  =  (26/5)

(vii) [x/(x+1)] + [(x + 1)/x] = 34/15

(viii) a²b²x² – (a² - b²) x + 1 = 0

(ix) 2(x + 1)² – 5 (x + 1)  =  12

(x) 3 (x – 4)² – 5(x – 4)  =  12

Solution

Question 1 :

3x²–5x–12  =  0

Solution :

3x²–5x–12  =  0

3x²–9x+4x–12  =  0

3x(x–3)+4(x–3)  =  0

(3x+4) (x-3)  =  0

So, the solution is {-4/3, 3}.

Question 2 :

(2x + 3)²-81  =  0

Solution :

(2x + 3)²-81  =  0

(2x + 3)² =  81

Taking square roots on both sides, we get

(2x + 3) =  √81

(2x + 3) =  ±9

So, the solution is {-6, 3}.

Question 3 :

√5x²+2x–3√5  =  0

Solution :

 √5x²+5x-3x–3√5  =  0

√5x(x+√5)–3(x+√5)  =  0

(√5x–3)(x+√5)  =  0

So, the solution is {-√5, 3/√5}.

Question 4 :

3(x²–6)  =  x(x+7)–3

Solution :

3(x²–6)  =  x(x+7)–3

3x²–18  =  x²+7x–3

3x²-x²–7x–18+3  =  0

2x²–7x–15  =  0

2x²–10x+3x–15  =  0

2x(x–5)+3(x–5)  =  0

(2x+3)(x–5)  =  0

So, the solution is {-3/2, 5}.

Question 5 :

3x – (8/x)  =  2

Solution :

(3x² – 8)/x  =  2

(3x² – 8)  =  2x

3x²-2x–8  =  0

3x²-6x+4x–8  =  0

3x(x–2)+4(x–2)  =  0

(3x+4) (x–2)  =  0

So, the solution is {-4/3, 2}.

Question 6 :

x+(1/x)  =  (26/5)

Solution :

(x² + 1)/x  =  26/5

5(x²+1)  =  26x

5x²+5  =  26x

5x²-26x+5  =  0

5x²-25x–x+5  =  0

5x(x-5)–1(x–5)  =  0

(5x–1) (x–5)  =  0

So, the solution is {1/5, 5}.

Question 7 :

[x/(x+1)]+[(x+1)/x]  =  34/15

Solution :

[x/(x+1)]+[(x+1)/x]  =  34/15

[x²+(x+1)²]/[x(x+1)]  =  34/15

[x²+(x²+2x+1)]/(x²+x)  =  34/15

15 (2x²+2x+1)  =  34 (x²+x)

30x²+30x+15  =  34x²+34x

34x²-30x²+34x–30x–15  =  0

4x²+4x–15  =  0

4x²+10x-6x–15  =  0

2x(2x+5)–3(2x+5)  =  0

(2x–3)(2x+5)  =  0

So, the solution is {-5/2, 3/2}.

Question 8 :

a²b²x²– (a²-b²)x+1  =  0

Solution :

a²b²x²– (a²-b²)x+1  =  0

a²b²x²– a²x+b²x+1  =  0

a²x (b²x–1)-1(b²x–1)  =  0

(b²x–1) (a²x–1)  =  0

So, the solution is {1/a², 1/b²}.

Question 9 :

2(x+1)² – 5(x+1)  =  12

Solution :

Let y  =  x+1

2y²–5y  =  12

2y²–5y–12  =  0

2y²–8y+3y–12  =  0

2y(y–4)+3(y–4)  =  0

(2y+3) (y–4)  =  0

Now we have to apply the values of y in the equation

y  =  x + 1

So, the solution is {-5/2, 3}.

Question 10 :

3(x–4)²– 5(x – 4)  =  12

Solution :

Let y = x – 4

3y²–5y  =  12

3y²–5y–12  =  0

3y²–9y+4y-12  =  0

3y(y–3)+4(y–3)  =  0

(3y+4)(y–3)  =  0

To find the value of x we have to apply the values of y in the equation x  =  y + 4.

So, the solution is {8/3, 7}.

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