Question

1) If one roots of x² + px + q = 0 is the square of other. Prove that p³ + q² = 3pq
2) ABCD is a cyclic quadrilateral. If <BCD = 30°, <ABC = 100°. Find<ACB​

Answer 1

Solution refer to the attachment........

hope it helps.....

Answer 2

1.) Correct Question:-

If one root of x² + px + q = is the square of other. Prove that p³ + q² + q = 3pq

1) Given:-

• p(x) = x² + px + q = 0
• One root of p(x) is the square of other.

To Proof:-

p³ + q²+q = 3pq

Assumption:-

Let the two zeroes of x² + px + q be $\sf{\alpha}$ and $\sf{\beta}$

Solution:-

We know,

$\sf{Sum\:of\:zeroes = \dfrac{-Coefficient\:of\:x}{Coefficient\:of\:x^2}}$

= $\sf{\alpha + \beta = \dfrac{-p}{1}}$

= $\sf{\alpha + \beta = -p \longrightarrow[i]}$

$\sf{Product\:of\:zeroes = \dfrac{Constant\:term}{Coefficient\:of\:x^2}}$

= $\sf{\alpha\beta = \dfrac{q}{1}}$

= $\sf{\alpha\beta = q \longrightarrow[ii]}$

Also given that,

One of the roots is square of the other.

Hence,

$\sf{\alpha = \beta^2}$

Substituting the value of $\sf{\underline{\alpha}}$ In eq.[i] and [ii].

Eq.[i]

= $\sf{\alpha + \beta = -p}$

= $\sf{\beta^2 + \beta = -p}$

=> $\sf{p = -(\beta^2 + \beta)}$

=> $\sf{p = -\beta^2 - \beta}$

Eq.[ii]

= $\sf{\alpha\beta = q}$

= $\sf{\beta^2\times \beta = q}$

= $\sf{\beta^3 = q}$

=> $\sf{q = \beta^3}$

Now,

$\sf{p^3 + q^2+ q = 3pq}$

LHS,

$\sf{(-\beta^2 -\beta)^3 + (\beta^3)^2+ \beta^3}$

= $\sf{-(\beta^2 + \beta)^3 + (\beta^6)}$

Using the identity:-

$\sf{(a+b)^3 = a^3 +3a^2b + 3ab^2 + b^3}$

= $\sf{-[(\beta^2)^3 + 3\times(\beta^2)^2\times\beta + 3\times\beta^2\times(\beta)^2+(\beta)^3]+\beta^6+\beta^3}$

= $\sf{-[\beta^6 + 3\beta^5 + 3\beta^4+\beta^3]+\beta^6+\beta^3}$

= $\sf{-\beta^6 - 3\beta^5 - 3\beta^4 -\beta^3 + \beta^6 + \beta^3}$

= $\sf{\cancel{-\beta^6}} - 3\beta^5 -3\beta^4 \cancel{-\beta^3} + \cancel{\beta^6}+{\cancel{\beta^3}}$

= $\sf{-3\beta^5-3\beta^4}$

RHS,

$\sf{3pq}$

= $\sf{3(-\beta-\beta^2)\beta^3}$

= $\sf{-3\beta^4 - 3\beta^5}$

=> $\sf{-3\beta^5-3\beta^4}$

Therefore LHS = RHS (Proved)

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2) Given:-

• ABCD is a cyclic quadrilateral.
• $\sf{\angle BCD = 30^\circ}$
• $\sf{\angle ABC = 100^\circ}$

To find:-

$\sf{\angle ACB}$

Note:-

Refer to the attachment for the diagram of the quadrilateral.

Solution:-

Since the measures of angles given are triangles

We know that,

$\sf{Sum \:of\: a \:triangle = 180^\circ}$

Therefore,

$\sf{\angle BCD +\angle ABC +\angle ACB = 180^\circ}$

By substituting,

$\sf{30^\circ + 100^\circ + \angle ACB = 180^\circ}$

$\sf{130^\circ + \angle ACB = 180^\circ}$

By transposing $\sf{130^\circ}$

$\sf{\angle ACB = 180^\circ - 130^\circ}$

$\sf{\angle ACB = 50^\circ}$

$\sf{\therefore \angle ACB \:measures\: 50^\circ}$

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