Question

question 42 please! It's quadratic equations.

Answer 1

$Given : \sqrt{x + 15} = x + 3$

On squaring both sides, we get

= > (x + 15) = (x + 3)^2

= > x + 15 = x^2 + 9 + 6x

= > x^2 + 6x - 6 = x

= > x^2 + 5x - 6 = 0

On comparing with ax^2 + bx + c = 0, we get a = 1, b = 5, c = -6

(1)

$= > x = \frac{-b + \sqrt{b^2 - 4ac}}{2a}$

$= > \frac{(-5) + \sqrt{(5)^2 - 4(1)(-6)} }{2}$

$= > \frac{-5 + \sqrt{25 + 24}}{2}$

$= > \frac{-5 + \sqrt{49}}{2}$

$= > \frac{-5 + 7}{2}$

= > 2/2

= > 1.

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(2)

$= > x = \frac{-b - \sqrt{b^2 - 4ac}}{2a}$

$= > \frac{5 - \sqrt{(5)^2 - 4(1)(-6)}}{2}$

$= > \frac{-5 - \sqrt{49}}{2}$

$= > \frac{-5 - 7}{2}$

= > -12/2

= > -6

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Verification:

When x = 1:

$= > \sqrt{1 + 15} = 1 + 3$

$= > \sqrt{16} = 4$

= > 4 = 4

when x = -6:

$= > \sqrt{-6 + 15} = -6 + 3$

$= > \sqrt{9} = -3$

= > 3 = -3( False).

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Therefore, the value of x = 1.

Hope this helps!