Question

If alpha and beta r zero of the polynomial f(x)=kx²+4x+4 such that alpha²+ beta²=24. Find the required value of k.





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Answer 1

f(x) = kx^2 + 4x +4

alpha + beta = -4/k

alpha × beta = 4/k

alpha^2 + beta^2 = 24

( alpha+ beta)^2 - 2 ( alpha)( beta) = 24

16/k^2 - 8/k - 24 = 0

3k^2 + k - 2 = 0

3k^2 + 3k -2k -2 = 0

3k ( k +1) -2( k +1) = 0

( 3k -2)( k +1)= 0

k= 2/3, -1

Answer 2

Solution:

Let $\bold{\alpha}$and $\bold{\beta}$be the zeros of the quadratic polynomial.

We know that a quadratic polynomial is of the form ax²+bx+c.

f(x)= kx²+4x+4

${\star}{\boxed{ \bold{ \underline{ \alpha + \beta = \frac{ - 4}{k} }}}}$

and,

$\star \boxed{ \bold{ \underline{\alpha \beta = \frac{4}{k} }}}$

Now,

$\boxed{ \bold{ \underline{\alpha {}^{2} + \beta {}^{2} = 24}}}$

$\bold { \implies ( \alpha + \beta ) {}^{2} - 2 \alpha \beta = 24}$

$\bold{ \implies( \frac{ - 4}{k} ) {}^{2} - 2 \times \frac{4}{k} = 24}$

$\bold {\implies \frac{16}{k {}^{2} } - \frac{8}{k} = 24 }$

$\bold{ \implies \: 16 - 8k = 24k {}^{2} }$

$\bold{ \implies \: 3k {}^{2} + k - 2 = 0 }$

$\bold { \implies3k {}^{2} + 3k - 2k - 2 = 0}$

$\bold{ \implies3k(k + 1) - 2(k + 1 = 0)}$

$\bold{ \implies \:( k + 1)(3 k- 2) = 0}$

$\boxed{\bold{ \implies \: k + 1 = 0 \: \: \:or \: \: 3k - 2 = 0}}$

$\huge{ \purple \star}{\boxed{ \red{ \mathfrak{ \underline {k = - 1 \: \: or \: \: k = \frac{ 2}{3} }}}}}$