A) 7 |
B) 1 |
C) \( \Large \frac{50}{7} \) |
D) \( \Large \frac{7}{50} \) |
A) 7 |
Given equation is \( \Large 7x^{2}-50x+k=0 \)
Here, a = 7, b = -50, c = k
Since, \( \Large \alpha + \beta =\frac{-b}{a} \)
Therefore, \( \Large \alpha + \beta =\frac{50}{7} \)
or \( \Large \beta =\frac{50}{7}-7 \)
=> \( \Large \beta =\frac{1}{7} \) [Because \( \Large \alpha = 7 \)(given)]
and \( \Large \alpha \beta =\frac{c}{a} \)
or \( \Large 7 \times \frac{1}{7}=\frac{k}{7} \)
k = 7
1). Find the roots of the equation \( \Large 2x^{2}-11x+15=0 \)
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2). The quadratic equation whose roots are 3 and -1, is
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3). \( \Large x^{2}+x-20=0; y^{2}-y-30=0 \)
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4). \( \Large 225x^{2}-4=0; \sqrt{225y}+2=0 \)
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5). \( \Large \frac{4}{\sqrt{x}}+\frac{7}{\sqrt{x}}=\sqrt{x;} \) \( \Large y^{2}-\frac{ \left(11\right)^{\frac{5}{2}} }{\sqrt{y}} =0 \)
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6). \( \Large x^{2}-365=364; y-\sqrt{324}=\sqrt{81} \)
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7). \( \Large 3x^{2}+8x+4=0; 4y^{2}-19y+12=0 \)
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8). \( \Large x^{2}-x-12=0; y^{2}+5y+6=0 \)
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9). \( \Large x^{2}-8x+15=0; y^{2}-3y+2=0 \)
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10). \( \Large x^{2}-32=112; y-\sqrt{169}=0 \)
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