The formula for finding roots of a quadratic equation a(x)2 + bx +c = 0 (derivation given in section 3.6.2) is x = (−b ± √[b2 − 4ac]) / 2a

**Solving
a Quadratic Equation by Formula Method**

The formula for finding
roots of a quadratic equation *ax*^{2} + *bx* +*c* = 0
(derivation given in section 3.6.2) is

Solve** ***x*^{2}** **+** **2*x*** **−** **2** **=** **0** **by formula method

Compare* **x*^{2}** **+

a = 1, b = 2, c = -2

substituting the values
of *a*, *b* and *c* in the formula we get,

Therefore, x = −1 + √3 , −1 − √3

Solve** **2*x*^{2} −
3*x* − 3 = 0 by formula method.

Compare** **2

substituting the values
of *a*, *b* and *c* in the formula we get,

**Example 3.35 **Solve** **3*p*^{2} +
2√5*p* – 5 = 0 by formula method

*Solution *

Solution Compare 3*p*^{2}
+ 2√5*p* – 5 = 0 with the Standard form
*ax*^{2} + *bx* +c = 0

a = 3, b = 2√5, c = −5 .

p = (−b ± √[b^{2} − 4ac]) / 2a

substituting the values of a, b and c in the
formula we get,

x = √5/3 , - √5

Solve** ***pqx*** **^{2}** **−** **(** ***p*** **+*q*)^{2}** ***x*** **+** **(** ***p*** **+*q*)^{2}** **=** **0

*Solution*

Compare the coefficients
of the given equation with the standard form *ax *^{2}* *+*
bx *+*c *=* *0

*a *= *pq* , *b* = −(*p* +*q*)^{2} , c = ( *p* +*q*)^{2}

substituting the values of *a*, *b* and *c* in the formula we get,

Tags : Example, Solution | Algebra Example, Solution | Algebra

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