Question

answer with solution​

Answer 1

121.

The option that satisfies the given condition is (1) 2b²=9ac

Explanation:

Let, One Quadratic equation, whose one root is double of the other.

E.g. (x+1)(x+2)=x²+3x+2

Here, 2(b)²=9(ac)

2×(3²)=9(1)(2)

18=18

E.g. (x+2)(x+4)=x²+6x+8

Here, 2b²=9ac

2(6)²=9(1)(8)

72=72

Hence, proved.

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122.

Answer: (3) b/ac

Explanation:

Let one root be (py) then other is (y).

Sum of roots=–b/a=py+y

$\frac{ - b}{ay} = p + 1$

$(p + 1) {}^{2} = ( - \frac{b}{ay} ) {}^{2} = \frac{b}{a {}^{2}.y {}^{2} }$

Product of Roots=c/a=py²

$p = \frac{c}{a.y {}^{2} }$

Then,

$\frac{(p + 1) {}^{2} }{p} = \frac{ \frac{b}{a {}^{2}y {}^{2} } }{ \frac{c}{ay {}^{2} } } = \frac{b(ay {}^{2}) }{c(a {}^{2} {y}^{2} )} = \frac{b}{ac}$

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123.

Answer: (1) q/p²

Let one root be (y), then other is (ky);

Sum of roots=ky+y=–b/a = p

» (k+1)=p/y

Product of roots=ky²=c/a=q

» k=q/y²

$∴ \frac{k}{(k + 1) {}^{2} } = \frac{q}{y {}^{2} } \div ( \frac{p}{y} ) {}^{2} = \frac{q}{p {}^{2} }$

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124.

Answer: (1) p²+p–2q=0

Let, one root be 'ɑ' and other be 'ß'

Given: ɑ+ß=ɑ²+ß²

Sum of roots, ɑ+ß=–p

Product of roots, ɑß=q

Sum of squares of roots:

ɑ²+ß²=(ɑ+ß)²–2ɑß [(a+b)²=a²+2ab+b²]

»ɑ²+ß²=(–p)²–2(q)

=p²–2q

As, ɑ+ß=ɑ²+ß²

» (–p)=(p²–2q)

» p²+p–2q=0

Answer 2

Answer:

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Step-by-step explanation:

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