Question

Solve for x 9x^2 – 15 = x^2 + 6

Answer 1

Given Equation is 9x^2 - 15x + 6 = 0.

(i)

Divide throughout by 9.

⇒ x^2 - (15/9) x + (6/9) = 0

⇒ x^2 - (5/3) x + (2/3) = 0

(ii)

Rewrite the equation with the constant term on the right side.

⇒ x^2 - (5/3) x = -2/3

(iii)

Complete the square by adding the square of one-half of coefficient.

⇒ x^2 - (5/3)x  + (5/6)^2 = -2/3 + (5/6)^2

⇒ x^2 - (5/3) x + (5/6)^2 = 1/36

(iv)

Write the left side as square.

⇒ (x - 5/6)^2 = 1/36

⇒ x - 5/6 = √1/36

⇒ x - 5/6 = 1/6

(v)

Equate and solve.

(a)

⇒ x - 5/6 = 1/6

⇒ x = 1/6 + 5/6

⇒ x = 1.

(b)

⇒ x - 5/6 = -1/6

⇒ x = -1/6 + 5/6

⇒ x = 2/3.

Therefore, roots are 1 and 2/3.

Hope it helps!

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Answer 2

$\huge{\bold{ Heya\: Dost}}$

$9 {x}^{2} - 15 = {x}^{2} + 6 \\ \\ 9 {x}^{2} - {x}^{2} = 6 + 15 \\ \\ 8 {x}^{2} = 19 \\ \\ {x}^{2} = \frac{19}{8} \\ \\ x = \sqrt{ \frac{19}{8} } \\ \\ x = \frac{ \sqrt{19} }{ \sqrt{2 \times 2 \times 2} } \\ \\ x = \frac{ \sqrt{19} }{2 \sqrt{2} } \\ \\ x = \frac{ \sqrt{19} \times \sqrt{2} }{2 \sqrt{2} \times \sqrt{2} } \\ \\ x = \frac{ \sqrt{38} }{4}$

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