3+√7/3-√7=a+b√7 solve step by step
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Answered by
2
3+√7 × 3+√7 =[3+√7]² = 3²+2×3×√7+(√7)² 9+6√7+7 16+6√7 16 √7
------- -------------------- ------------------------ = -----------------= -------------= -------+ ----
3-√7 × 3+√7 =3²-(√7)² = 9-7 2 2 2 2
= 8+3√7
a+b√7
a=8;b=3
------- -------------------- ------------------------ = -----------------= -------------= -------+ ----
3-√7 × 3+√7 =3²-(√7)² = 9-7 2 2 2 2
= 8+3√7
a+b√7
a=8;b=3
Answered by
15
ⓗⓔⓨ ⓕⓡⓘⓔⓝⓓ
ⓗⓔⓡⓔ ⓘⓢ ⓨⓞⓤⓡ
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![\frac{3 + \sqrt{7} }{3 - \sqrt{7} } = a + b \sqrt{7} \frac{3 + \sqrt{7} }{3 - \sqrt{7} } = a + b \sqrt{7}](https://tex.z-dn.net/?f=+%5Cfrac%7B3+%2B+%5Csqrt%7B7%7D+%7D%7B3+-+%5Csqrt%7B7%7D+%7D+%3D+a+%2B+b+%5Csqrt%7B7%7D+)
solving by rationalising the denominator:-
![\frac{3 + \sqrt{7} }{3 - \sqrt{7} } \times \frac{3 + \sqrt{7} }{3 + \sqrt{7} } = \frac{(3 + \sqrt{7)} {}^{2} }{3 {}^{2} - (\sqrt{7}) {}^{2} } \frac{3 + \sqrt{7} }{3 - \sqrt{7} } \times \frac{3 + \sqrt{7} }{3 + \sqrt{7} } = \frac{(3 + \sqrt{7)} {}^{2} }{3 {}^{2} - (\sqrt{7}) {}^{2} }](https://tex.z-dn.net/?f=+%5Cfrac%7B3+%2B+%5Csqrt%7B7%7D+%7D%7B3+-+%5Csqrt%7B7%7D+%7D+%5Ctimes+%5Cfrac%7B3+%2B+%5Csqrt%7B7%7D+%7D%7B3+%2B+%5Csqrt%7B7%7D+%7D+%3D+%5Cfrac%7B%283+%2B+%5Csqrt%7B7%29%7D+%7B%7D%5E%7B2%7D+%7D%7B3+%7B%7D%5E%7B2%7D+-+%28%5Csqrt%7B7%7D%29+%7B%7D%5E%7B2%7D+%7D+)
=>
![\frac{16 + 6 \sqrt{7} }{2} = a + b \sqrt{7} \frac{16 + 6 \sqrt{7} }{2} = a + b \sqrt{7}](https://tex.z-dn.net/?f=+%5Cfrac%7B16+%2B+6+%5Csqrt%7B7%7D+%7D%7B2%7D+%3D+a+%2B+b+%5Csqrt%7B7%7D+)
=> so,
![a = \frac{16}{2 } = 8 \\ \\ b = \frac{6}{2} = 3 a = \frac{16}{2 } = 8 \\ \\ b = \frac{6}{2} = 3](https://tex.z-dn.net/?f=a+%3D+%5Cfrac%7B16%7D%7B2+%7D+%3D+8+%5C%5C+%5C%5C+b+%3D+%5Cfrac%7B6%7D%7B2%7D+%3D+3)
![<b > hope it helps you <b > hope it helps you](https://tex.z-dn.net/?f=+%26lt%3Bb+%26gt%3B+hope+it+helps+you)
@ Prabhudutt
ⓗⓔⓡⓔ ⓘⓢ ⓨⓞⓤⓡ
ⓐⓝⓢⓦⓔⓡ
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solving by rationalising the denominator:-
=>
=> so,
@ Prabhudutt
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hii
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