Question

(1) Solve given equations by elimination method 3x + y = 5 and 5x - y = 11.
(2) Verify whether 3 is the zero of P(x) = 2x3 - 11x2 + 17x - 6
(3) If a, ß are zeroes of ax2 - 6x-
6 and aß= 4, find the value of a.plz solve on paper​

Answer 1

(1.)

GIVEN :–

• 3x + y = 5 & 5x - y = 11

TO FIND :–

• Value of 'x' & 'y' = ?

SOLUTION :–

▪︎ 3x + y = 5 ______________eq.(1)

▪︎ 5x - y = 11 ______________eq.(2)

• Add eq.(1) & eq.(2) –

=> (3x + y) + (5x - y) = 5 + 11

=> 8x = 16

=> x = 2

• Using eq.(2) –

=> 5(2) - y = 11

=> 10 - y = 11

=> y = -1

▪︎ Hence the value of x = 2 & value of y = -1 .

(2.)

• Zero's of polynomial always satisfy the Polynomial .

=> p(3) = 0

• Let's take L.H.S. –

= p(3)

= 2(3)³ - 11(3)² + 17(3) -6

= 2(27) - 11(9) + 51 - 6

= 54 - 99 + 51 -6

= 105 - 105

= 0

= R.H.S.

▪︎ Hence, '3' is the zero of given polynomial.

(3.)

GIVEN :–

• A quadratic equation ax² - 6x - 6 = 0.

• Product of roots (αβ ) = 4

TO FIND :–

• Value of 'a' = ?

SOLUTION :–

• We know that –

• Product of roots = αβ = [constant term]/[coffieciant of x²]

=> αβ = (-6)/a

=> 4 = (-6)/a

=> 4a = -6

=> a = -6/4

=> a = -3/2

Answer 2

$\bf \huge \blue{answer}$

$\bf \bold \: 1)solution :$

$\bf \: given :$

$\bf \: 3x + y = 5...........(1) \\ \bf \: 5x - y = 11..........(2)$

$\rm \bold \: subtract \: eq \: (1) \: and \: (2)$

$\bf \: 3x \: + y \: = \: 5 \\ \bf \underline{ \: 5x \: - y \: = \: 11 } \\ \bf \: 8x \: \: \: \: \: \: \: \: \: \: = \: 16 \\ \bf \: x = \frac{16}{8} \\ \bf \: x = 2$

x=2 substitute in eq(1)

$\bf \: 3x + y = 5 \\ \bf \: 3(2) + y = 5 \\ \bf \: 6 + y = 5 \\ \bf \: y = 5 - 6 \\ \bf \: y = - 1$

$\bf \huge\fbox \green{ x = 2 \: \: and \: \: y = - 1}$

$\rm \huge{ \underline {\underline {\red{2) \: verification}}}}$

$\bf {\: p(3) = 2 {(3)}^{3} - 11 {(3)}^{2} + 17(3) - 6} \\ \bf = 2(27) - 11(9) + 51 - 6 \\ \bf \: = 54 - 99 + 51 - 6 \\ \bf \: = 105 - 105 \\ \bf \: = 0$

$\bf \: hence \: verified \: \\ \bf \: 3 \: is \: the \: zeroes \: of \: the \: polynomial$

$\bf \huge \purple{3)solution}$

$\bf \: product \: of \: zeroes \: = \alpha \beta = 4 \\ \bf \: \alpha \beta = 4 = \frac{c}{a} \\ \bf \: 4 = \frac{ - 6}{a } \\ \bf \: 4a = - 6 \\ \bf \: a = \frac{ - 6}{4} \\ \bf \: a = \frac{ - 3}{2}$