Question

hey plz answer this
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Answer 1

Required Answer:-

Given:

• $\rm \log_{10}(98 + \sqrt{ {x}^{2} - 12x + 36 } ) = 2$

To Find:

• The values of x.

Solution:

We have,

$\rm \implies\log_{10}(98 + \sqrt{ {x}^{2} - 12x + 36 } ) = 2$

$\rm \implies 98 + \sqrt{ {x}^{2} - 12x + 36 } = {10}^{2}$

$\rm \implies \sqrt{ {x}^{2} - 12x + 36 } = 100 - 98$

$\rm \implies \sqrt{ {x}^{2} - 12x + 36 } = 2$

Squaring both sides, we get,

$\rm \implies {x}^{2} - 12x + 36 = 4$

$\rm \implies {x}^{2} - 12x + 36 - 4 = 0$

$\rm \implies {x}^{2} - 12x + 32= 0$

Splitting the middle term,

$\rm \implies {x}^{2} - 4x - 8x + 32= 0$

$\rm \implies x(x - 4) - 8(x - 4)= 0$

$\rm \implies (x - 8)(x - 4)= 0$

By zero product rule,

➡ Either (x - 8) = 0 or (x - 4) = 0

$\rm \implies x = 4,8$

★ Hence, the roots of this equation are 4 and 8.

Answer:

• The roots are - 4 and 8 (Option B)

Formula Used:

• log(m) = n ➡ m = 10ⁿ

Answer 2

Answer is option D

Hope it helps you:)