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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
[tex] \frac{ - b}{ay} = p + 1[/tex]
[tex](p + 1) {}^{2} = ( - \frac{b}{ay} ) {}^{2} = \frac{b}{a {}^{2}.y {}^{2} } [/tex]
Product of Roots=c/a=py²
[tex]p = \frac{c}{a.y {}^{2} } [/tex]
Then,
[tex] \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} [/tex]
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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
[tex]∴ \frac{k}{(k + 1) {}^{2} } = \frac{q}{y {}^{2} } \div ( \frac{p}{y} ) {}^{2} = \frac{q}{p {}^{2} } [/tex]
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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:
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