CBSE > Class 10 > Mathematics | CBSE Board | Homework Help

Let {tex}\alpha,\beta{/tex} be the zeros of the polynomial {tex}f(x)=x^2-8x+k{/tex}.
Sum of zeroes = {tex} \alpha + \beta = - \left( \frac { - 8 } { 1 } \right) = 8{/tex} and, Product of zeroes = {tex} \alpha \beta = \frac { k } { 1 } = k{/tex}
Now,
{tex} \alpha ^ { 2 } + \beta ^ { 2 } = 40{/tex}

{tex} \Rightarrow \alpha ^ { 2 } + \beta ^ { 2 }+2 \alpha\beta-2 \alpha\beta= 40{/tex}
{tex} \Rightarrow \quad ( \alpha + \beta ) ^ { 2 } - 2 \alpha \beta = 40{/tex}
{tex} \Rightarrow \quad 8 ^ { 2 } - 2 k = 40{/tex}
{tex} \Rightarrow \quad 2 k = 64 - 40 {/tex}

{tex}\Rightarrow 2 k = 24 {/tex}

{tex}\Rightarrow k = 12{/tex}

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ANSWER

X2_2x_8=0

Posted by Hariom Dhakar 7 years, 6 months ago

CBSE > Class 10 > Mathematics

0 answers

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If h is the hcf of 609 and 957.find x and y, satisfy h=609x+957y?

Posted by Harshveer Natt 7 years, 6 months ago

CBSE > Class 10 > Mathematics

0 answers

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[sin30°+tan45°-cosec60°]÷[sec30°+cos60°+cot45°]

Posted by Kuldeep Solanki 7 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Aditya Singh 7 years, 6 months ago

Put the values and divide

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Trigonometry means

Posted by Tushar Mhaske 7 years, 6 months ago

CBSE > Class 10 > Mathematics

3 answers

Amogh Joshi 7 years, 6 months ago

Measure of Three angels of triangles

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Divanshu Rawat 7 years, 6 months ago

Knoledge of three angles

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Nishant Kumar 7 years, 6 months ago

Dimaag Ka dahi -(दिमाग का दही)

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if sin thetha + cos thetha = √2 then the value of tan25 thetha + cot25 thetha + sec2 thetha + cos2 thetha is??

Posted by Nikhil Sharma 7 years, 6 months ago

CBSE > Class 10 > Mathematics

0 answers

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55x + 67y =311 67 x + 55y = 299

Posted by Hritik Tak 7 years, 6 months ago

CBSE > Class 10 > Mathematics

4 answers

Pragya . 7 years, 6 months ago

Ayush Mishra

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Pragya . 7 years, 6 months ago

i don't think that your answer is correct plz check

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Pragya . 7 years, 6 months ago

by substitution method 55x +67y =311 ----------eq. 1 67x +55y =299 ----------eq. 2 from eq. 1 we obtain x = 311 - 67y / 55 ----------eq. 3 substituting the value of x in eq. 2 67(311 - 67y / 55) + 55y = 299 (20837 - 4489y /55) + 55y =299 20837 - 4489y + 3025y = 299 * 55 20837 - 1464y = 16445 -1464y = 16445 - 20837 -1464y = -4392 y = -4392 / -1464 y = 3 putting the value of y in eq. 3 x = 311 - 67(3) / 55 x = 311 - 201 / 55 x = 110/55 x = 2 x = 2 and y = 3

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Ayush Mishra 7 years, 6 months ago

x = 4 and y = 3

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ANSWER

Find the zeroes of the polynomials and verify the relationship between the zeroes and coefficient of xsquare-8x+15

Posted by Nischal Rajineesh 7 years, 6 months ago

CBSE > Class 10 > Mathematics

0 answers

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If one zero of tje polynomial (a^2+9)x^2+13x+6a is reciprocal of the other find value of a.

Posted by Yashraj Thakur 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

Let {tex} \alpha{/tex} and {tex} \frac { 1 } { \alpha }{/tex} be the zeros of (a²+ 9)x²+ 13x + 6a.
Then, we have
{tex} \alpha \times \frac { 1 } { \alpha } = \frac { 6 a } { a ^ { 2 } + 9 }{/tex}
⇒ 1 = {tex} \frac { 6 a } { a ^ { 2 } + 9 }{/tex}
⇒ a² + 9 = 6a
⇒ a² - 6a + 9 = 0
⇒ a² - 3a - 3a + 9 = 0
⇒ a(a - 3) - 3(a - 3) = 0
⇒ (a - 3) (a - 3) = 0
⇒ (a - 3)²= 0
⇒ a - 3 = 0
⇒ a = 3
So, the value of a in given polynomial is 3.