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x² + 5√5x - 70 = 0, find whether the equation has real roots. If real roots exist, find them

Solution:

Given, the equation is x² + 5√5x - 70 = 0

We have to find whether the equation has real roots or not.

A quadratic equation ax² + bx + c = 0 has 2 distinct real roots when the discriminant of the equation is greater than zero.

Discriminant = b² - 4ac

Here, a = 1, b = 5√5 and c = -70

So, b² - 4ac = (5√5)² - 4(1)(-70)

= 125 + 280

= 405 > 0

So, the equation has 2 distinct real roots.

By using the quadratic formula,

x = [-b ± √b² - 4ac]/2a

x = (-5√5 ± √405)/2(1)

= (-5√5 ± 9√5)/2

Now, x = (-5√5 + 9√5)/2 = 4√5/2 = 2√5

x = (-5√5 - 9√5)/2 = -14√5/2 = -7√5

Therefore, the roots of the equation are -7√5 and 2√5.

✦ Try This: Check whether the equation x² + 5√5x - 3 = 0 has real roots. If real roots exist, find them

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 4

NCERT Exemplar Class 10 Maths Exercise 4.4 Problem 1 (v)

Summary:

The equation x² + 5√5x - 70 = 0 has real roots which can be written as x = 2√5 and x = -7√5

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