Question

If root over 13-p root 10 = root8 + root 5 then find p.​

Answer 1

Solution

$\rm \sqrt{13 - p \sqrt{10} } = \sqrt{8} + \sqrt{5}$

Squaring on both sides

$\implies \bigg(\sqrt{13 - p \sqrt{10} } \bigg) ^{2} = (\sqrt{8} + \sqrt{5} )^{2}$

$\implies 13 - p \sqrt{10} = { (\sqrt{8}) }^{2} + {( \sqrt{5} )}^{2} + 2( \sqrt{8} )( \sqrt{5} )$

[ Because (a + b)² = a² + b² + 2ab ]

$\implies 13 - p \sqrt{10} = 8 + 5 + 2 \sqrt{8 \times 5}$

$\implies 13 - p \sqrt{10} = 8 + 5 + 2 \sqrt{40}$

$\implies 13 - p \sqrt{10} = 8 + 5 + 2 \sqrt{4 \times 10}$

$\implies 13 - p \sqrt{10} = 13 + 4\sqrt{10}$

$\implies 13 - 13 - 4 \sqrt{10} = p \sqrt{10}$

$\implies - 4 \sqrt{10} = p \sqrt{10}$

$\implies \dfrac{ - 4 \sqrt{10} }{ \sqrt{10} } = p$

$\implies - 4 = p$

$\implies \boxed{p = - 4}$

Hence the value of p is - 4.

Answer 2

Question :----

$\sqrt{13 - p \sqrt{10} } = \sqrt{8} + \sqrt{5}$

Find the value of P ?

Formula used :---

• (a+b)² = (a² + b² + 2ab)

Solution :----

Squaring both sides of the given Equation we get,

since (√a)² = a ,

in LHS we have = (13 - P√10)

in RHS we have = (√8 + √5)²

so,

13 - P√10 = 8 + 5 + 2√40

→ 13 - P√10 = 13 + 2√(4×10)

→ 13 - P√10 = 13 + 2×2√10

→ 13 - P√10 = 13 + 4√10 .

Comparing here we get, P = (-4)

or we can solve it further , since 13 cancel out both sides ,

we get,

→ (-P√10) = 4√10

(√10) also cancel out now ,

→ (-P) = 4

or,

→ (-1)P = 4

→ P = (-4) (Ans)

(Hope it helps you)