Find the value of a and b- 7 - -5-7 - -5 - 7 - -5-7 - -5 - a - 7-11-5 bHence factorise the polynomial - 24x2 - a - 41x - b - 11

The value of 7+√(5)7 √(5) 7 √(5)7+√(5)=(7+√(5))^2 (7 √(5))^27^2 (√(5))^2=711√(5) Hence a=0 and b=1, substitute the same in the polynomial 24x^2 41x+12 ...

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Byju's Answer

Standard X

Mathematics

Quadratic Formula

Find the valu...

Question

Find the value of a and b.
$\frac{7 + \sqrt{5}}{7 - \sqrt{5}} - \frac{7 - \sqrt{5}}{7 + \sqrt{5}} = a + \frac{7}{11} \sqrt{5} b$
Hence factorise the polynomial : $24 x^{2} - (a + 41) x + (b + 11)$

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Solution

The value of $\frac{7 + \sqrt{5}}{7 - \sqrt{5}} - \frac{7 - \sqrt{5}}{7 + \sqrt{5}} = \frac{(7 + \sqrt{5})^{2} - (7 - \sqrt{5})^{2}}{7^{2} - (\sqrt{5})^{2}} = \frac{7}{11} \sqrt{5}$
Hence a=0 and b=1, substitute the same in the polynomial
$24 x^{2} - 41 x + 12$
$24 x^{2} - 32 x - 9 x + 12$
$8 x (3 x - 4) - 3 (3 x - 4)$
$\Rightarrow (8 x - 3) (3 x - 4)$

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Similar questions

Q. Find the values of a and b if

\frac{7 + \sqrt{5}}{7 - \sqrt{5}} - \frac{7 - \sqrt{5}}{7 + \sqrt{5}} = a + \frac{7}{11} \sqrt{5} b

Q. If

\frac{7 + \sqrt{5}}{7 - \sqrt{5}} - \frac{7 - \sqrt{5}}{7 + \sqrt{5}} = a + \frac{7 \sqrt{5}}{11}

b then find value of a and b.

Q. When denominator is rationalised, then the number

\frac{7 + \sqrt{5}}{7 - \sqrt{5}} - \frac{7 - \sqrt{5}}{7 + \sqrt{5}}

becomes

a + \frac{7}{11} \sqrt{5} b

. Find

a + b

Q. Question 11
Find the values of a and b in each of the following

(i) \frac{5 + 2 \sqrt{3}}{7 + 4 \sqrt{3}} = a - 6 \sqrt{3}

(i i) \frac{3 - \sqrt{5}}{3 + 2 \sqrt{5}} = a \sqrt{5} - \frac{19}{11}

(i i i) \frac{\sqrt{2} + \sqrt{3}}{3 \sqrt{2} - 2 \sqrt{3}} = 2 - b \sqrt{6}

(i v) \frac{7 + \sqrt{5}}{7 - \sqrt{5}} - \frac{7 - \sqrt{5}}{7 + \sqrt{5}} = a + \frac{7}{11} \sqrt{5} b

Q. Find the values of a and b:

\frac{7 + \sqrt{5}}{7 - \sqrt{5}} - \frac{7 - \sqrt{5}}{7 + \sqrt{5}} = a + 7 \sqrt{5} b

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Find the value of a and b.7+√57−√5−7−√57+√5=a+711√5bHence factorise the polynomial : 24x2−(a+41)x+(b+11)

The value of 7+√57−√5−7−√57+√5=(7+√5)2−(7−√5)272−(√5)2=711√5Hence a=0 and b=1, substitute the same in the polynomial24x2−41x+1224x2−32x−9x+128x(3x−4)−3(3x−4)⇒(8x−3)(3x−4)

Find the value of a and b.
$\frac{7 + \sqrt{5}}{7 - \sqrt{5}} - \frac{7 - \sqrt{5}}{7 + \sqrt{5}} = a + \frac{7}{11} \sqrt{5} b$
Hence factorise the polynomial : $24 x^{2} - (a + 41) x + (b + 11)$