ML Aggarwal Solution Class 10 Chapter 5 Quadratic Equations in One Variable Exercise 5.1
Exercise 5.1
Question 1
Check whether the following are quadratic equations:
It is a quadratic equation as it is power of 2.
(ii)
(2x + 1) (3x – 2) = 6(x + 1) (x – 2)
6x² – 4x + 3x – 2 = 6(x² – 2x + x – 2)
6x² – x – 2 = 6x² – 12x + 6x – 12
12x-6x-x=-12+2
5x=-10
It is not a quadratic equation.
(iii)
(x−3)3+5=x3+7x2−1
x3−3x2×3+3x×9−27+5=x3+7x2−1
−9x2+27x−22−7x2+1=0
−16x2+27x−21=0
16x2−27x+21=0
It is a quadratic equation
(iv)
It is a quadratic equation
(v)
x+2x=x2,x≠0
It is not a quadratic equation
(vi)
Question 2
In each of the following, determine whether the given numbers are roots of the given equations or not;
(i) x² – x + 1 = 0; 1, – 1
(ii) x² – 5x + 6 = 0; 2, – 3
(iii) 3x² – 13x – 10 = 0; 5,−23
(iv) 6x² – x – 2 = 0; −12,23
Sol :
(i)
x² – x + 1 = 0; 1, -1
Where x = 1, then
(1)² – 1 + 1 = 1 – 1 + 1 = 1 ≠ 0
∴ x = 1 does not satisfy it
∴x=-1, does not satisfy it
∴x=1,-1 are not roots of the equation
(ii)
x2−5x+6=0;2,−3
When x=2, then
(2)2−5×2+6
=4-10+6=10-10=0
∴x=2 is its root
Where x=-3, then
(−3)2−5(−3)+6
=9+15+6=30≠0
∴x=-3 is not its solution
∴2 is root of the equation by -3 is not a root.
(iii)
3x2−13x−10=0;5,−23
x=5
3(5)2−13×5−10=75−65−10
=75-75=0
∴x=5 is it root
If x=−23, then
3(−23)2−13×−23−10
=3×49+263−10
=43+263−10=303−10=10−10=0
∴x=−23 is also its root.
Hence both 5,−23 are its roots.
(iv)
6x2−x−2=0;−12,23
If x=−12, then
=6(−12)2−(−12)−2
=6×14+12−2=32+12−2
=42−2=0
∴x=−12 is its root
If x=23, then
=6×49−23−2=83−23−2
=63−2=0
∴x=23 is also its root.
Hence −12,23 are both its root.
Question 3
In each of the following, determine whether the given numbers are solutions of the given equation or not:
(i) x² – 3√3x + 6 = 0; √3, – 2√3
(ii) x² – √2x – 4 = 0, x = – √2, 2√2
Sol :
(i) x² – 3√3x + 6 = 0; √3, -2√3
(a) Substituting the value of x = √3
=(√3)2−3√3×√3+6=3−9+6=0=R.H.S
∴x=√3 is its solution
(b) x=−2√3
Substituting x=−2√3
L.H.S. =x2−3√3x+6
=(−2√3)2−3√3(−2√3)+6
=12+18+6=36≠0
∵x=−2√3 is not its solution
(ii) x2−√2x−4=0,x=−√2,2√2
(a) x=−√2
Substituting x=−√2
L.H.S.=x2−√2x−4
=(−√2)2−√2(−√2)−4
=2+2-4=0=R.H.S
∴x=−√2 is its solution
(b) x=−2√2
Substituting x=−2√2
L.H.S. =x2−√2x−4
=(−2√2)2−√2(−2√2)−4
=8-4-4=8-8=0=R.H.S
∴x=−2√2 is its solution
Question 4
(i) If −12 is a solution of the equation 3x² + 2kx – 3 = 0, find the value of k.
Sol :
x=−12 is a solution of the
3x² + 2kx – 3 = 0,
Substituting the value of x in the given equation
3×14−k−3=0
34−k−3=0
⇒k=34−3=−94
Hence k=−94
(ii) 7x2+kx−3=0,x=23
∵x=23 is its solution
∴7(23)2+k(23)−3=0
⇒7×49+23k−3=0
⇒289−3+23k=0
⇒23k=3−289
⇒23k=27−289⇒23k=−19
⇒k=−19×32=−16
Hence k=−16
Question 5
(i) If √2 is a root of the equation kx² + √2 – 4 = 0, find the value of k.
(ii) If a is a root of the equation x² – (a + b)x + k = 0, find the value of k.
Sol :
(i) kx² + √2 – 4 = 0, x = √2
x = √2 is its solution
Question 6
If 23 and – 3 are the roots of the equation px² + 7x + q = 0, find the values of p and q.
Substituting the value of x=23 and – 3 respectively, we get
⇒49p+143+q=0
⇒4 p+42+9 q=0
⇒4p+9q=-42...(i)
and p(−3)2+7(−3)+q=0
⇒9p-21+q=0
⇒q=21-9p...(ii)
Substituting the value of q in (i)
⇒4 p+9(21-9 p)=-42
⇒4 p+189-81 p=-42
⇒-77 p=-42-189=-231
⇒p=−231−77=3
∴q=21-9×3
=21-27=-6
∴p=3, q=-6
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