If α and β(α>β) are the roots of the equation [1 5][ 1 3−4 7]2[719 −319419 119]4[ 1 3−4 7]2[x2−5x+5 −3]=[−4] then the value of (2+α)(2−β)

Let A=[ 1 3−4 7] then A−1=[719 −319419 119]Since A−1A=I=AA−1,[ 1 3−4 7][719 −319419 119]=I=[719 −319419 119][ 1 3−4 7] [1 5] [x2−5x+5 −3]=[−4]⇒x2−5x+5−15=−4⇒x2−5x−6=0⇒x=6,−1(α,β)

If α and β(α>β) are the roots of the equation [1 5][ 1 3−4 7]2[719 −319419 119]4[ 1 3−4 7]2[x2−5x+5 −3]=[−4] then the value of (2+α)(2−β)

Let A=[ 1 3−4 7] then A−1=[719 −319419 119]Since A−1A=I=AA−1,[ 1 3−4 7][719 −319419 119]=I=[719 −319419 119][ 1 3−4 7] [1 5] [x2−5x+5 −3]=[−4]⇒x2−5x+5−15=−4⇒x2−5x−6=0⇒x=6,−1(α,β)

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If $α$ and $β (α > β)$ are the roots of the equation $[1 5] {[\begin{array}{l} 1 3 \\ - 4 7 \end{array}]}^{2} {[\begin{array}{l} \frac{7}{19} - \frac{3}{19} \\ \frac{4}{19} \frac{1}{19} \end{array}]}^{4} {[\begin{array}{l} 1 3 \\ - 4 7 \end{array}]}^{2} [\begin{array}{l} x^{2} - 5 x + 5 \\ - 3 \end{array}] = [- 4]$ then the value of $(2 + α) (2 - β)$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$- 12$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Let $A = [\begin{array}{l} 1 3 \\ - 4 7 \end{array}]$ then $A^{- 1} = [\begin{array}{l} \frac{7}{19} - \frac{3}{19} \\ \frac{4}{19} \frac{1}{19} \end{array}]$
Since $A^{- 1} A = I = A A^{- 1}, [\begin{array}{l} 1 3 \\ - 4 7 \end{array}] [\begin{array}{l} \frac{7}{19} - \frac{3}{19} \\ \frac{4}{19} \frac{1}{19} \end{array}] = I = [\begin{array}{l} \frac{7}{19} - \frac{3}{19} \\ \frac{4}{19} \frac{1}{19} \end{array}] [\begin{array}{l} 1 3 \\ - 4 7 \end{array}]$
$[1 5] [\begin{array}{l} x^{2} - 5 x + 5 \\ - 3 \end{array}] = [- 4] \Rightarrow x^{2} - 5 x + 5 - 15 = - 4 \Rightarrow x^{2} - 5 x - 6 = 0 \Rightarrow x = 6, - 1 (α, β)$