Question

Please help with part (b) of this question!! thank you!!!​

Answer 1

`\textsf{(a)} \quad h=3, \quad k=-2`

`\textsf{(b)} \quad p=0, \quad q=-(1)/(2)`

Explanation:

Part (a)

Rearrange the given equation so that both sides are in the same form:

`\begin{aligned}2-h√(11)&=-3√(11)-k\\2-h√(11)+k&=-3√(11)-k+k\\k+2-h√(11)&=-3√(11)\\k+2-h√(11)-2&=-3√(11)-2\\k-h√(11)&=-2-3√(11)\\\\\implies h&=3\\\implies k&=-2\end{aligned}`

Part (b)

Rewrite and factor the right side of the equation.

As p and q are rationals, both sides of the equation have to equal zero.

`\begin{aligned}2p&=q√(76)+√(19)\\2p&=q√(4 \cdot 19)+√(19)\\2p&=q√(4) √(19)+√(19)\\2p&=2q√(19)+√(19)\\2p&=√(19)(2q+1)\\\\\implies 2p&=0 \implies p=0\\\implies 2q+1&=0 \implies q=-(1)/(2)\end{aligned}`

Answer 2

• p = 0, q = - 0.5

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Part (b)

Simplify first:

• 2p = q√76 + √19
• 2p = q(√4×√19) + √19
• 2p = 2q√19 + √19
• 2p = √19(2q + 1)

We need both p and q be rational numbers and we know √19 is irrational.

It is only possible if both side are equal to zero:

• 2p = 0 ⇒ p = 0,
• 2q + 1 = 0 ⇒ 2q = - 1 ⇒ q = - 0.5