Question

solve using sum and product rule ?
$\sqrt{2 } x {}^{2} - 3x - 2 \sqrt{2} = 0$
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Answer 1

Question:

Solve the given equation using sum and product rule ; √2x² - 3x - 2√2 = 0 .

Answer:

2√2 , -√2/2

Note:

∆ The general form of a quadratic equation is given as ; ax² + bx + c = 0.

∆ If A and B are the zeros of the quadratic equation ax² + bx + c = 0 , then ;

• Sum of roots,(A+B) = - b/a

• Product of roots,(A•B) = c/a

∆ (x+y)² = x² + y² + 2•x•y

∆ (x-y)² = x² + y² - 2•x•y

∆ (x+y)•(x-y) = x² - y²

∆ (x-y)² = (x+y)² - 4•x•y

∆ (x+y)² = (x-y)² + 4•x•y

Solution:

Here,

The given quadratic equation is ;

√2x² - 3x - 2√2 = 0 .

Clearly,

a = √2

b = -3

c = -2√2

Let A and B be the roots of given quadratic equation.

Thus ;

The sum of the roots will be ;

=> A + B = -b/a

=> A + B = -(-3)/√2

=> A + B = 3/√2 --------(1)

Also,

The product of the roots will be ;

=> A•B = c/a

=> A•B = -2√2/√2

=> A•B = -2 -------(2)

Now,

=> (A-B)² = (A+B)² - 4•A•B

=> (A-B)² = (3/√2)² - 4•(-2) { using eq-(1) and (2) }

=> (A-B)² = 9/2 + 8

=> (A-B)² = (9+16)/2

=> (A-B)² = 25/2

=> A - B = √(25/2)

=> A - B = 5/√2 --------(3)

Now,

Adding eq-(1) and eq-(3) , we have ;

=> A + B + A - B = 3/√2 + 5/√2

=> 2A = (3+5)/√2

=> 2A = 8/√2

=> A = 8/2√2

=> A = 2√2

Now ,

Putting A = 2√2 in eq-(1) , we get ;

=> A + B = 3/√2

=> A + B = 3√2/2

=> 2√2 + B = 3√2/2

=> B = 3√2/2 - 2√2

=> B = (3√2 - 4√2)/2

=> B = -√2/2

Hence,

The required roots of given quadratic equation are : 2√2 and -√2/2 .