Question

any five question answers​

Answer 1

Since −5 is a root of the equation 2x

Since −5 is a root of the equation 2x 2

Since −5 is a root of the equation 2x 2 +px−15=0. Therefore,

Since −5 is a root of the equation 2x 2 +px−15=0. Therefore,2(−5)

Since −5 is a root of the equation 2x 2 +px−15=0. Therefore,2(−5) 2

Since −5 is a root of the equation 2x 2 +px−15=0. Therefore,2(−5) 2 −5p−15=0⇒p=7

Since −5 is a root of the equation 2x 2 +px−15=0. Therefore,2(−5) 2 −5p−15=0⇒p=7Putting p=7 in p(x

Since −5 is a root of the equation 2x 2 +px−15=0. Therefore,2(−5) 2 −5p−15=0⇒p=7Putting p=7 in p(x 2

Since −5 is a root of the equation 2x 2 +px−15=0. Therefore,2(−5) 2 −5p−15=0⇒p=7Putting p=7 in p(x 2 +x)+k=0 we get

Since −5 is a root of the equation 2x 2 +px−15=0. Therefore,2(−5) 2 −5p−15=0⇒p=7Putting p=7 in p(x 2 +x)+k=0 we get7x

Since −5 is a root of the equation 2x 2 +px−15=0. Therefore,2(−5) 2 −5p−15=0⇒p=7Putting p=7 in p(x 2 +x)+k=0 we get7x 2

Since −5 is a root of the equation 2x 2 +px−15=0. Therefore,2(−5) 2 −5p−15=0⇒p=7Putting p=7 in p(x 2 +x)+k=0 we get7x 2 +7x+k=0

Since −5 is a root of the equation 2x 2 +px−15=0. Therefore,2(−5) 2 −5p−15=0⇒p=7Putting p=7 in p(x 2 +x)+k=0 we get7x 2 +7x+k=0Here ,a=7,b=7;c=k

Since −5 is a root of the equation 2x 2 +px−15=0. Therefore,2(−5) 2 −5p−15=0⇒p=7Putting p=7 in p(x 2 +x)+k=0 we get7x 2 +7x+k=0Here ,a=7,b=7;c=kThis equation will have equal roots, if

Since −5 is a root of the equation 2x 2 +px−15=0. Therefore,2(−5) 2 −5p−15=0⇒p=7Putting p=7 in p(x 2 +x)+k=0 we get7x 2 +7x+k=0Here ,a=7,b=7;c=kThis equation will have equal roots, ifDiscriminant =b

Since −5 is a root of the equation 2x 2 +px−15=0. Therefore,2(−5) 2 −5p−15=0⇒p=7Putting p=7 in p(x 2 +x)+k=0 we get7x 2 +7x+k=0Here ,a=7,b=7;c=kThis equation will have equal roots, ifDiscriminant =b 2

Since −5 is a root of the equation 2x 2 +px−15=0. Therefore,2(−5) 2 −5p−15=0⇒p=7Putting p=7 in p(x 2 +x)+k=0 we get7x 2 +7x+k=0Here ,a=7,b=7;c=kThis equation will have equal roots, ifDiscriminant =b 2 −4ac=0 ⇒49−4×7×k=0⇒k=

Since −5 is a root of the equation 2x 2 +px−15=0. Therefore,2(−5) 2 −5p−15=0⇒p=7Putting p=7 in p(x 2 +x)+k=0 we get7x 2 +7x+k=0Here ,a=7,b=7;c=kThis equation will have equal roots, ifDiscriminant =b 2 −4ac=0 ⇒49−4×7×k=0⇒k= 28

Since −5 is a root of the equation 2x 2 +px−15=0. Therefore,2(−5) 2 −5p−15=0⇒p=7Putting p=7 in p(x 2 +x)+k=0 we get7x 2 +7x+k=0Here ,a=7,b=7;c=kThis equation will have equal roots, ifDiscriminant =b 2 −4ac=0 ⇒49−4×7×k=0⇒k= 2849

Since −5 is a root of the equation 2x 2 +px−15=0. Therefore,2(−5) 2 −5p−15=0⇒p=7Putting p=7 in p(x 2 +x)+k=0 we get7x 2 +7x+k=0Here ,a=7,b=7;c=kThis equation will have equal roots, ifDiscriminant =b 2 −4ac=0 ⇒49−4×7×k=0⇒k= 2849

Since −5 is a root of the equation 2x 2 +px−15=0. Therefore,2(−5) 2 −5p−15=0⇒p=7Putting p=7 in p(x 2 +x)+k=0 we get7x 2 +7x+k=0Here ,a=7,b=7;c=kThis equation will have equal roots, ifDiscriminant =b 2 −4ac=0 ⇒49−4×7×k=0⇒k= 2849 ⇒k=

Since −5 is a root of the equation 2x 2 +px−15=0. Therefore,2(−5) 2 −5p−15=0⇒p=7Putting p=7 in p(x 2 +x)+k=0 we get7x 2 +7x+k=0Here ,a=7,b=7;c=kThis equation will have equal roots, ifDiscriminant =b 2 −4ac=0 ⇒49−4×7×k=0⇒k= 2849 ⇒k= 4

Since −5 is a root of the equation 2x 2 +px−15=0. Therefore,2(−5) 2 −5p−15=0⇒p=7Putting p=7 in p(x 2 +x)+k=0 we get7x 2 +7x+k=0Here ,a=7,b=7;c=kThis equation will have equal roots, ifDiscriminant =b 2 −4ac=0 ⇒49−4×7×k=0⇒k= 2849 ⇒k= 47

Since −5 is a root of the equation 2x 2 +px−15=0. Therefore,2(−5) 2 −5p−15=0⇒p=7Putting p=7 in p(x 2 +x)+k=0 we get7x 2 +7x+k=0Here ,a=7,b=7;c=kThis equation will have equal roots, ifDiscriminant =b 2 −4ac=0 ⇒49−4×7×k=0⇒k= 2849 ⇒k= 47

Since −5 is a root of the equation 2x 2 +px−15=0. Therefore,2(−5) 2 −5p−15=0⇒p=7Putting p=7 in p(x 2 +x)+k=0 we get7x 2 +7x+k=0Here ,a=7,b=7;c=kThis equation will have equal roots, ifDiscriminant =b 2 −4ac=0 ⇒49−4×7×k=0⇒k= 2849 ⇒k= 47

Since −5 is a root of the equation 2x 2 +px−15=0. Therefore,2(−5) 2 −5p−15=0⇒p=7Putting p=7 in p(x 2 +x)+k=0 we get7x 2 +7x+k=0Here ,a=7,b=7;c=kThis equation will have equal roots, ifDiscriminant =b 2 −4ac=0 ⇒49−4×7×k=0⇒k= 2849 ⇒k= 47 Solve any question of Quadratic Equations with:-