Question

5x square minus px+1=0 and alpha minus beta=1 then find p

Answer 1

Appropriate Question :-

$\sf \: \alpha , \beta \: are \: roots \: of \: {5x}^{2} - px + 1 = 0 \: and \: \alpha - \beta = 1, \: find \: p.$

$\large\underline{\sf{Solution-}}$

Given that,

$\sf \: \alpha , \beta \: are \: roots \: of \: {5x}^{2} - px + 1 = 0$

We know,

$\boxed{\red{\sf Sum\ of\ the\ roots=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\rm\implies \: \alpha + \beta = - \dfrac{( - p)}{5} = \dfrac{p}{5}$

Also,

$\boxed{\red{\sf Product\ of\ the\ roots=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\sf\implies \: \alpha \beta = \dfrac{1}{5}$

Now, Given that,

$\sf \: \alpha - \beta = 1$

On squaring both sides, we get

$\sf \: (\alpha - \beta)^{2} = 1$

$\sf \: { \alpha }^{2} + { \beta }^{2} - 2 \alpha \beta = 1$

$\sf \: { \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta - 4 \alpha \beta = 1$

$\sf \: {(\alpha + \beta )}^{2} - 4 \alpha \beta = 1$

$\sf \: {\bigg(\dfrac{p}{5} \bigg) }^{2} - 4 \times \dfrac{1}{5} = 1$

$\sf \: \dfrac{ {p}^{2} }{25} - \dfrac{4}{5} = 1$

$\sf \: \dfrac{ {p}^{2} }{25} = \dfrac{4}{5} + 1$

$\sf \: \dfrac{ {p}^{2} }{25} = \dfrac{4 + 5}{5}$

$\sf \: \dfrac{ {p}^{2} }{25} = \dfrac{9}{5}$

$\sf \: {p}^{2} = 45$

$\sf \: p \: = \: \pm \: \sqrt{45}$

$\sf \: p \: = \: \pm \: \sqrt{3 \times 3 \times 5}$

$\rm\implies \:\sf \: p \: = \: \pm \: 3\sqrt{5}$

$\rule{190pt}{2pt}$

Additional Information :-

$\boxed{ \rm{ \: { \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta )}^{2} - 2 \alpha \beta \: }}$

$\boxed{ \rm{ \: { \alpha }^{3} + { \beta }^{3} = {( \alpha + \beta )}^{3} - 3 \alpha \beta ( \alpha + \beta ) \: }}$

$\boxed{ \rm{ \: \alpha - \beta = \sqrt{ {( \alpha + \beta) }^{2} - 4 \alpha \beta } \: }}$

$\boxed{ \rm{ \: { \alpha }^{4} + { \beta }^{4} = {[( { \alpha + \beta) }^{2} - 2 \alpha \beta ]}^{2} - 2 {( \alpha \beta )}^{2} \: }}$

Answer 2

5x²- px + 1 = 0 and α - β = 1 then, find p.

You must have studied the two special concepts for roots of Quadratic Equation (Trust me you can handle 90 percent of Quadratic question using that.. rest 10 can be solved by Additional concepts of Quadratic Fraction, graphs and all.)

The Form of Quadratic equation ax² + bx + c = 0.

let, α & β be two roots of this equation

Sum of the roots = -b/a = - (-p)/5 = p/5

Product of the roots = c/a = 1/5

Just see, α - β = 1 => (1)

We have to use some algebraic identity where we can substitute the value for αβ and α+β.

(α - β)² = 1²

α² + β² - 2αβ = 1

Now, see we can't implement the value for α+β until we bring the eqń (α+β)²

(α² + β² + 2αβ) - 4αβ = 1

* equation remains constant.

(α+β)² - 4αβ = 1 (*claps* we're close to the ans)

=> (p/5)² - 4/5 = 1

=> (p²/25) - 4/5 = 1

=> p²/25 = 9/5

=> p²/5 = 9

p² = 45

p = ± 3√5