If-the-equation-4x-2-4-5x-1-p-2-0-has-one-root-equals-to-two-more-then-the-other-then-the-value-of-p-is-equal-to-

Question Number 101056 by bemath last updated on 30/Jun/20

If the equation 4 x^{2} - 4 (5 x + 1) + p^{2} = 0

has one root equals to two more

then the other, then the value of

p is equal to___

Commented by bemath last updated on 30/Jun/20

thank you both

Answered by Rasheed.Sindhi last updated on 30/Jun/20

Still another way

4 x^{2} - 20 x + p^{2} - 4 = 0 \dots \dots . (i)

T h e e q u a t i o n w i t h α & α + 2 r o o t s

(x - α) (x - α - 2) = 0

x^{2} - (2 α + 2) x + α^{2} + 2 α = 0 \dots . (i i)

C o m p a r i n g c o e f f i c i e n t s o f (i) & (i i)

\frac{4}{1} = \frac{- 20}{- (2 α + 2)} = \frac{p^{2} - 4}{α^{2} + 2 α}

8 α + 8 = 20 \land p^{2} - 4 = 4 α^{2} + 8 α

α = \frac{3}{2} \land p^{2} = 4 {(\frac{3}{2})}^{2} + 8 (\frac{3}{2}) + 4

p^{2} = 9 + 12 + 4 = 25

p = \pm 5

Commented by bemath last updated on 30/Jun/20

waw \dots great sir . thank you

Answered by Rasheed.Sindhi last updated on 30/Jun/20

4 x^{2} - 4 (5 x + 1) + p^{2} = 0

4 x^{2} - 20 x - 4 + p^{2} = 0

α + (α + 2) = \frac{- (- 20)}{4} \land α (α + 2) = \frac{p^{2} - 4}{4}

2 α + 2 = 5 \Rightarrow α = \frac{3}{2}

\Rightarrow \frac{p^{2} - 4}{4} = α (α + 2) = \frac{3}{2} (\frac{3}{2} + 2)

\Rightarrow \frac{p^{2} - 4}{4} = \frac{3}{2} (\frac{7}{2})

\Rightarrow p^{2} = 21 + 4 \Rightarrow p = \pm 5

Answered by Ar Brandon last updated on 30/Jun/20

Let the roots be α and α + 2 . Then;

Sum of roots 2 α + 2 = \frac{20}{4} = 5 \Rightarrow α = \frac{3}{2}

Product of roots α (α + 2) = α^{2} + 2 α = \frac{p^{2} - 4}{4}

\Rightarrow {(\frac{3}{2})}^{2} + 2 (\frac{3}{2}) = \frac{p^{2} - 4}{4} = \frac{21}{4} \Rightarrow p = \pm 5

Answered by Rasheed.Sindhi last updated on 30/Jun/20

AnOther Way

4 x^{2} - 20 x + p^{2} - 4 = 0

α i s a r o o t (s a y)

4 α^{2} - 20 α + p^{2} - 4 = 0 \dots \dots . (i)

α + 2 i s o t h e r r o o t

4 {(α + 2)}^{2} - 20 (α + 2) + p^{2} - 4 = 0

4 α^{2} + 16 α + 16 - 20 α - 40 + p^{2} - 4 = 0

4 α^{2} - 4 α + p^{2} - 28 = 0 \dots \dots . (i i)

(i) - (i i) : - 16 α + 24 = 0 \Rightarrow α = \frac{3}{2}

N o w,

(i) \Rightarrow 4 {(\frac{3}{2})}^{2} - 20 (\frac{3}{2}) + p^{2} - 4 = 0

9 - 30 + p^{2} - 4 = 0 \Rightarrow p = \pm 5

Answered by Rasheed.Sindhi last updated on 30/Jun/20

One way more : A simple way

(W h e t h e r y o u l i k e i t o r d i s l i k e,

a n y w a y t h i s i s a l s o a w a y .)

4 x^{2} - 20 x + p^{2} - 4 = 0

x = \frac{5 \pm \sqrt{29 - p^{2}}}{2}

\frac{5 + \sqrt{29 - p^{2}}}{2} - \frac{5 - \sqrt{29 - p^{2}}}{2} = 2

\frac{5 + \sqrt{29 - p^{2}} - 5 + \sqrt{29 - p^{2}}}{2} = 2

\sqrt{29 - p^{2}} = 2

29 - p^{2} = 4

p^{2} = 25

p = \pm 5

Commented by john santu last updated on 01/Jul/20

cooll

Commented by Rasheed.Sindhi last updated on 01/Jul/20

T h a n k s S i r!