Question

Find the root of the equation (x+3) (x-3)=25

Answer 1

Step-by-step explanation:

isolate one of the square roots:√(2x−5) = 1 + √(x−1) square both sides:2x−5 = (1 + √(x−1))2 ...

expand right hand side:2x−5 = 1 + 2√(x−1) + (x−1) ...

isolate the square root:√(x−1) = (x−5)/2. ...

Expand right hand side:x−1 = (x2 − 10x + 25)/4. ...

Multiply by 4 to remove division:4x−4 = x2 − 10x + 25.

Answer 2

Given : -

( x + 3 ) ( x - 3 ) = 25

Required to find : -

• The roots of the Equation ?

Formula used : -

$\boxed{\sf{\bf{ x = \dfrac{ - b \pm \sqrt{ b^2 - 4ac } }{ 2a } } }}$

Solution : -

➾ ( x + 3 ) ( x - 3 ) = 25

Using the identity ;

( a + b ) ( a - b ) = a² - b²

➾ ( x + 3 ) ( x - 3 ) = 25

➾ ( x )² - ( 3 )² = 25

➾ x² - 9 = 25

➾ x² = 25 + 9

➾ x² = 34

➾ x² - 34 = 0

Hence,

The Quadratic Equation is x² - 34 = 0

Now,

Let's solve this quadratic equation to find the roots .

The standard form of the quadratic equation is

ax² + bx + c = 0

Compare the standard form with the above quadratic equation .

➾ x² - 34 = 0 & ax² + bx + c = 0

Here,

• a = 1

• b = 0

• c = - 34

Using the Quadratic formula ;

$\boxed{\sf{\bf{ x = \dfrac{ - b \pm \sqrt{ b^2 - 4ac } }{ 2a } } }}$

This implies ;

$\mathtt{ x = \dfrac{ - (0) \pm \sqrt{(0 {)}^{2} - 4(1)( - 34) } }{2(1)} } \\ \\ \\ \mathtt{x = \frac{0 \pm \sqrt{0 - 4( - 34)} }{2} } \\ \\ \\ \mathtt{x = \frac{0 \pm \sqrt{0 - ( - 136)} }{2} } \\ \\ \\ \mathtt{x = \frac{0 \pm \sqrt{0 + 136} }{2} } \\ \\ \\ \mathtt{x = \frac{0 \pm \sqrt{136} }{2} } \\ \\ \\ \mathtt{x = \frac{0 + \sqrt{4 \times 34} }{2} \quad \: ( \: and \: ) \quad x = \frac{0 - \sqrt{4 \times 34} }{2} } \\ \\ \\ \mathtt{x = \frac{0 + 2 \sqrt{34} }{2} \quad \: ( \: and \: ) \: \quad \: x = \frac{0 - 2 \sqrt{34} }{2} } \\ \\ \\ \tt{x = \frac{ 2\sqrt{34} }{2} \quad \: ( \: and \: ) \: \quad x = \frac{ - 2 \sqrt{34} }{2} } \\ \\ \\ \sf{x = \sqrt{34} \quad \: ( \: and \: ) \quad x = - \sqrt{34} }$

Hence,

➾ x = √34 & x = -√34

Therefore,

Roots of the quadratic equation is √34 & -√34

Additional Information : -

$\boxed{\sf{\bf{ x = \dfrac{ - b \pm \sqrt{ b^2 - 4ac } }{ 2a } } }}$

This formula is named as " Quadratic formula " .

However,

This can be written as :

$\boxed{\sf{\bf{ \alpha = \dfrac{ - b + \sqrt{ b^2 - 4ac } }{ 2a } } }}$

$\boxed{\sf{\bf{\beta = \dfrac{ - b - \sqrt{ b^2 - 4ac } }{ 2a } } }}$

Here,

alpha , beta are the zeroes of the quadratic equation .

b² - 4ac is known to be as Discriminate .

Since,

It is able to discriminate the natur of the roots .

Discriminate is represented by " D " .

The conditions are ;

• If D > 0 ( perfect square )

Roots are distinct & rational.

• If D > 0 ( not perfect square )

Roots are distinct & irrational .

• If D = 0

Roots are equal and rational .

• If D < 0

Roots are distinct & imaginary .