Trying to find the right answer to this..

We Given -5 is a root of the quadratic equation 2x 2 +px-15 =0 , -5 satisfies the given equation. ∴ 2 (-5) 2 +p(-5)-15=0 50-5p-15 =0 35-5p=0 5p=35 ⇒ p=7 Substituting p=7 in (x 2 +x)+k =0, we get 7(x 2 +x)+k =0 7x 2 +7x+k=0 The roots of the equation are equal ∴ Discriminant =b 2 -4ac =0 Here, a= 7, b=7,c=k b 2 -4ac=0 ∴(7) 2 -4(7)(k) =0 49-28k =0 28k=49 k= 49/28 =7/4

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Quadratic Equations

Multiple Choice Questions

341. If one root of the quadratic equation 2x² – 3x + P = 0 is 3, then the other root is

$3 over 2$
$negative 3 over 2$
$1 half$
$1 half$

422 Views

342. If you divide 29 into two parts so that the sum of the squares of the parts is 425 then the two parts are

13, 15
16, 15
13, 16
13, 16

197 Views

343. For what value of a the quadratic equation (a - 4)x² + 2(α – 4)x + 4 = 0, has equal roots.

α = 7
α = 5
α = 4
α = 4

98 Views

344. The two consecutive positive integers, sum of whose square is 365 are

13, 14
12, 13
11, 12
11, 12

163 Views

345. One of the root of a quadratic equation

space square root of 2 straight x squared plus 7 straight x plus 5 square root of 2 space equals space 0

$space fraction numerator negative 5 over denominator square root of 2 end fraction$
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109 Views

346. The discriminant of the quadratic equation

3 straight x squared minus 4 square root of 3 straight x plus 4 equals 0

D > 0
D < 0
D = 0
D = 0

137 Views

Short Answer Type

347.
If -5 is a root of the quadratic equation 2x² + px – 15 = 0 and the quadratic equation p(x² + x)k = 0 has equal roots, find the value of k.

We Given -5 is a root of the quadratic equation 2x²+px-15 =0
, -5 satisfies the given equation.
∴ 2 (-5)² +p(-5)-15=0
50-5p-15 =0
35-5p=0
5p=35 ⇒ p=7

Substituting p=7 in (x²+x)+k =0, we get
7(x²+x)+k =0
7x²+7x+k=0
The roots of the equation are equal
∴ Discriminant =b²-4ac =0
Here, a= 7, b=7,c=k
b²-4ac=0
∴(7)²-4(7)(k) =0
49-28k =0
28k=49
k= 49/28
=7/4

3820 Views

348.

Solve for x:
$fraction numerator 1 over denominator left parenthesis straight x minus 1 right parenthesis left parenthesis straight x minus 2 right parenthesis end fraction plus fraction numerator 1 over denominator left parenthesis straight x minus 2 right parenthesis left parenthesis straight x minus 3 right parenthesis end fraction space equals 2 over 3 comma space straight x not equal to 1 comma 2 comma 3$

3303 Views

349.

Solve for x:

$fraction numerator 1 over denominator straight x plus 1 end fraction plus fraction numerator 2 over denominator x plus 2 end fraction plus fraction numerator 4 over denominator x plus 4 end fraction comma space x not equal to negative 1 comma negative 2 comma negative 4$